Step 1: Decomposition of the Function

Given the function $y = x^{\sin x} + (\sin x)^{\cos x}$. Let $u = x^{\sin x}$ and $v = (\sin x)^{\cos x}$. The function becomes $y = u + v$. To find the derivative $\frac{dy}{dx}$, we must compute $\frac{du}{dx} + \frac{dv}{dx}$ [By the linearity of the derivative operator].

Step 2: Differentiating $u = x^{\sin x}$

Taking the natural logarithm on both sides: $\ln(u) = \ln(x^{\sin x})$. Using the logarithmic property $\ln(a^b) = b \ln a$:

$\ln(u) = \sin x \cdot \ln x$

Differentiating both sides with respect to $x$ [Using the Chain Rule and Product Rule]:

$\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(\sin x) \cdot \ln x + \sin x \cdot \frac{d}{dx}(\ln x)$

$\frac{1}{u} \frac{du}{dx} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x}$

$\frac{du}{dx} = u \left( \cos x \ln x + \frac{\sin x}{x} \right) = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right)$

Step 3: Differentiating $v = (\sin x)^{\cos x}$

Taking the natural logarithm on both sides: $\ln(v) = \ln((\sin x)^{\cos x})$.

$\ln(v) = \cos x \cdot \ln(\sin x)$

Differentiating both sides with respect to $x$ [Using the Chain Rule and Product Rule]:

$\frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}(\cos x) \cdot \ln(\sin x) + \cos x \cdot \frac{d}{dx}(\ln(\sin x))$

Applying the Chain Rule to $\frac{d}{dx}(\ln(\sin x)) = \frac{1}{\sin x} \cdot \cos x = \cot x$:

$\frac{1}{v} \frac{dv}{dx} = -\sin x \cdot \ln(\sin x) + \cos x \cdot \cot x$

$\frac{dv}{dx} = v \left( \cos x \cot x - \sin x \ln(\sin x) \right) = (\sin x)^{\cos x} \left( \cos x \cot x - \sin x \ln(\sin x) \right)$

Step 4: Combining the Results

Since $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$, we aggregate the expressions obtained in Step 2 and Step 3:

$\frac{dy}{dx} = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right) + (\sin x)^{\cos x} \left( \cos x \cot x - \sin x \ln(\sin x) \right)$

Visual Representation of Derivative Decomposition

Final Solution: \frac{dy}{dx} = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right) + (\sin x)^{\cos x} \left( \cos x \cot x - \sin x \ln(\sin x) \right)

Step 1: Functional Representation

The given function is defined as:

$f(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)$

Step 2: Logarithmic Differentiation Setup

To differentiate the product of multiple terms efficiently, we apply the natural logarithm to both sides [Per the property of logarithms: $\ln(abc) = \ln a + \ln b + \ln c$]:

$\ln(f(x)) = \ln((1 + x)(1 + x^2)(1 + x^4)(1 + x^8))$

$\ln(f(x)) = \ln(1 + x) + \ln(1 + x^2) + \ln(1 + x^4) + \ln(1 + x^8)$

Step 3: Differentiation with Respect to $x$

Differentiating both sides with respect to $x$ [By the Chain Rule: $\frac{d}{dx}(\ln u) = \frac{1}{u} \cdot \frac{du}{dx}$]:

$\frac{1}{f(x)} \cdot f'(x) = \frac{d}{dx}[\ln(1 + x)] + \frac{d}{dx}[\ln(1 + x^2)] + \frac{d}{dx}[\ln(1 + x^4)] + \frac{d}{dx}[\ln(1 + x^8)]$

$\frac{f'(x)}{f(x)} = \frac{1}{1 + x} + \frac{2x}{1 + x^2} + \frac{4x^3}{1 + x^4} + \frac{8x^7}{1 + x^8}$

Step 4: Expression for $f'(x)$

Isolating $f'(x)$:

$f'(x) = f(x) \left( \frac{1}{1 + x} + \frac{2x}{1 + x^2} + \frac{4x^3}{1 + x^4} + \frac{8x^7}{1 + x^8} \right)$

Step 5: Evaluating at $x = 1$

First, calculate the value of $f(1)$:

$f(1) = (1 + 1)(1 + 1^2)(1 + 1^4)(1 + 1^8)$

$f(1) = (2)(2)(2)(2) = 16$

Now, substitute $x = 1$ into the derivative expression:

$f'(1) = f(1) \left( \frac{1}{1 + 1} + \frac{2(1)}{1 + 1^2} + \frac{4(1)^3}{1 + 1^4} + \frac{8(1)^7}{1 + 1^8} \right)$

$f'(1) = 16 \left( \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} \right)$

$f'(1) = 16 \left( \frac{1}{2} + 1 + 2 + 4 \right)$

$f'(1) = 16 \left( \frac{1 + 2 + 4 + 8}{2} \right)$

$f'(1) = 16 \left( \frac{15}{2} \right)$

$f'(1) = 8 \times 15 = 120$

Visual Verification of Function Trend

Final Solution: $f'(1) = 120$

Step 1: Formal Setup and Logarithmic Transformation

Let the given function be represented as $y = f(x)$:

$y = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} = \left[ \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \right]^{1/2}$

To differentiate this complex rational function, we apply the natural logarithm ($\ln$) to both sides to simplify the radical and quotient operations using logarithmic properties [Per logarithmic identity $\ln(a^b) = b\ln(a)$ and $\ln(ab/c) = \ln(a) + \ln(b) - \ln(c)$]:

$\ln y = \ln \left[ \left( \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \right)^{1/2} \right]$

$\ln y = \frac{1}{2} \left[ \ln((x-1)(x-2)) - \ln((x-3)(x-4)(x-5)) \right]$

$\ln y = \frac{1}{2} [ \ln(x-1) + \ln(x-2) - \ln(x-3) - \ln(x-4) - \ln(x-5) ]$

Step 2: Implicit Differentiation

Differentiating both sides with respect to $x$ using the chain rule [Per $\frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx}$]:

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{d}{dx}(\ln(x-1)) + \frac{d}{dx}(\ln(x-2)) - \frac{d}{dx}(\ln(x-3)) - \frac{d}{dx}(\ln(x-4)) - \frac{d}{dx}(\ln(x-5)) \right]$

Utilizing the standard derivative $\frac{d}{dx}(\ln(u)) = \frac{1}{u} \frac{du}{dx}$:

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right]$

Step 3: Solving for $\frac{dy}{dx}$

Multiply both sides by $y$ to isolate the derivative:

$\frac{dy}{dx} = \frac{y}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right]$

Substitute the original definition of $y$ back into the equation:

$\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right]$

Analytical Visualization

Final Solution: $\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \left( \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right)$

Step 1: Decomposition of the Function

Given the function $y = x^{x \cos x} + \frac{x^2+1}{x^2-1}$. To differentiate, we decompose the function into two parts, $u$ and $v$, such that $y = u + v$.

$u = x^{x \cos x}$

$v = \frac{x^2+1}{x^2-1}$

The derivative is given by the linearity property: $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ [Per the Sum Rule of Differentiation].

Step 2: Differentiating $u = x^{x \cos x}$

Since the variable is in both the base and the exponent, we apply logarithmic differentiation [By Properties of Logarithms: $\ln(a^b) = b \ln a$].

Taking the natural logarithm on both sides:

$\ln u = \ln(x^{x \cos x})$

$\ln u = (x \cos x) \cdot \ln x$

Differentiating with respect to $x$ on both sides [By Chain Rule and Product Rule]:

$\frac{1}{u} \frac{du}{dx} = \frac{d}{dx} [x \cos x] \cdot \ln x + (x \cos x) \cdot \frac{d}{dx} [\ln x]$

Applying the Product Rule again to $\frac{d}{dx}[x \cos x]$:

$\frac{1}{u} \frac{du}{dx} = [\cos x - x \sin x] \ln x + (x \cos x) \cdot \frac{1}{x}$

$\frac{du}{dx} = u \left( (\cos x - x \sin x) \ln x + \cos x \right)$

Substituting $u$ back into the expression:

$\frac{du}{dx} = x^{x \cos x} \left[ \cos x \ln x - x \sin x \ln x + \cos x \right]$

Step 3: Differentiating $v = \frac{x^2+1}{x^2-1}$

We use the Quotient Rule: $\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$.

Let $f(x) = x^2+1 \implies f'(x) = 2x$.

Let $g(x) = x^2-1 \implies g'(x) = 2x$.

$\frac{dv}{dx} = \frac{(2x)(x^2-1) - (x^2+1)(2x)}{(x^2-1)^2}$

$\frac{dv}{dx} = \frac{2x^3 - 2x - 2x^3 - 2x}{(x^2-1)^2}$

$\frac{dv}{dx} = \frac{-4x}{(x^2-1)^2}$

Step 4: Synthesis of the Final Derivative

Combining the derivatives obtained in Step 2 and Step 3:

$\frac{dy}{dx} = x^{x \cos x} (\cos x \ln x - x \sin x \ln x + \cos x) - \frac{4x}{(x^2-1)^2}$

Final Solution: $\frac{dy}{dx} = x^{x \cos x} (\cos x \ln x - x \sin x \ln x + \cos x) - \frac{4x}{(x^2-1)^2}$

Step 1: Defining the Function and Applying Logarithms

Let the given function be defined as:

$y = (x + 3)^2 \cdot (x + 4)^3 \cdot (x + 5)^4$

To simplify the differentiation of a product of multiple power functions, we apply the natural logarithm ($\ln$) to both sides. [Per the logarithmic property $\ln(abc) = \ln a + \ln b + \ln c$ and $\ln(a^n) = n \ln a$]:

$\ln(y) = \ln[(x + 3)^2 \cdot (x + 4)^3 \cdot (x + 5)^4]$

$\ln(y) = \ln(x + 3)^2 + \ln(x + 4)^3 + \ln(x + 5)^4$

$\ln(y) = 2\ln(x + 3) + 3\ln(x + 4) + 4\ln(x + 5)$

Step 2: Differentiating with Respect to $x$

Now, differentiate both sides with respect to $x$ using the Chain Rule [$\frac{d}{dx}(\ln u) = \frac{1}{u} \cdot \frac{du}{dx}$]:

$\frac{d}{dx}[\ln(y)] = \frac{d}{dx}[2\ln(x + 3) + 3\ln(x + 4) + 4\ln(x + 5)]$

$\frac{1}{y} \cdot \frac{dy}{dx} = 2 \cdot \frac{1}{x+3} + 3 \cdot \frac{1}{x+4} + 4 \cdot \frac{1}{x+5}$

Step 3: Isolating the Derivative $\frac{dy}{dx}$

Multiply both sides by $y$ to solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = y \cdot \left[ \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \right]$

Step 4: Substitution of the Original Function

Substitute the initial value of $y = (x + 3)^2 (x + 4)^3 (x + 5)^4$ into the equation:

$\frac{dy}{dx} = (x + 3)^2 (x + 4)^3 (x + 5)^4 \cdot \left[ \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \right]$

Step 5: Simplifying the Expression (Optional but Recommended)

We can distribute the factors inside the bracket for a more refined result:

$\frac{dy}{dx} = 2(x+3)(x+4)^3(x+5)^4 + 3(x+3)^2(x+4)^2(x+5)^4 + 4(x+3)^2(x+4)^3(x+5)^3$

Factoring out the common terms $(x+3)(x+4)^2(x+5)^3$:

$\frac{dy}{dx} = (x+3)(x+4)^2(x+5)^3 \cdot [2(x+4)(x+5) + 3(x+3)(x+5) + 4(x+3)(x+4)]$

Expanding the terms inside the square bracket:

$= (x+3)(x+4)^2(x+5)^3 \cdot [2(x^2 + 9x + 20) + 3(x^2 + 8x + 15) + 4(x^2 + 7x + 12)]$

$= (x+3)(x+4)^2(x+5)^3 \cdot [2x^2 + 18x + 40 + 3x^2 + 24x + 45 + 4x^2 + 28x + 48]$

$= (x+3)(x+4)^2(x+5)^3 \cdot [9x^2 + 70x + 133]$

Final Solution: $\frac{dy}{dx} = (x + 3)(x + 4)^2(x + 5)^3(9x^2 + 70x + 133)$

Step 1: Decomposition of the Implicit Equation

Given the implicit equation: $x^y + y^x = 1$.

To differentiate this with respect to $x$, we must define two separate functions, $u = x^y$ and $v = y^x$, such that:

$u + v = 1$

Differentiating with respect to $x$ yields:

$\frac{du}{dx} + \frac{dv}{dx} = 0$ [Per the Sum Rule of Differentiation]  --- (Equation 1)

Step 2: Logarithmic Differentiation of $u = x^y$

Taking the natural logarithm on both sides:

$\ln(u) = \ln(x^y) = y \ln(x)$ [By Logarithmic Power Rule]

Differentiating with respect to $x$ using the product rule:

$\frac{1}{u} \frac{du}{dx} = y \cdot \frac{d}{dx}(\ln x) + \ln x \cdot \frac{dy}{dx}$

$\frac{1}{u} \frac{du}{dx} = \frac{y}{x} + \ln x \frac{dy}{dx}$

$\frac{du}{dx} = x^y \left( \frac{y}{x} + \ln x \frac{dy}{dx} \right)$  --- (Equation 2)

Step 3: Logarithmic Differentiation of $v = y^x$

Taking the natural logarithm on both sides:

$\ln(v) = \ln(y^x) = x \ln(y)$

Differentiating with respect to $x$ using the product rule:

$\frac{1}{v} \frac{dv}{dx} = x \cdot \frac{d}{dx}(\ln y) + \ln y \cdot \frac{d}{dx}(x)$

$\frac{1}{v} \frac{dv}{dx} = x \cdot \frac{1}{y} \frac{dy}{dx} + \ln y \cdot (1)$

$\frac{dv}{dx} = y^x \left( \frac{x}{y} \frac{dy}{dx} + \ln y \right)$  --- (Equation 3)

Step 4: Algebraic Substitution and Solving for $\frac{dy}{dx}$

Substitute Equation 2 and Equation 3 into Equation 1:

$x^y \left( \frac{y}{x} + \ln x \frac{dy}{dx} \right) + y^x \left( \frac{x}{y} \frac{dy}{dx} + \ln y \right) = 0$

Expand the terms:

$x^y \cdot \frac{y}{x} + x^y \ln x \frac{dy}{dx} + y^x \frac{x}{y} \frac{dy}{dx} + y^x \ln y = 0$

Factor out $\frac{dy}{dx}$:

$\frac{dy}{dx} (x^y \ln x + x y^{x-1}) = -(y x^{y-1} + y^x \ln y)$

Isolate $\frac{dy}{dx}$:

$\frac{dy}{dx} = -\frac{y x^{y-1} + y^x \ln y}{x^y \ln x + x y^{x-1}}$

Summary Illustration

Final Solution: \frac{dy}{dx} = -\frac{y x^{y-1} + y^x \ln y}{x^y \ln x + x y^{x-1}}

Step 1: Decomposition of the Function

Let the given function be $y = f(x) + g(x)$, where:

• $f(x) = \left(x + \frac{1}{x}\right)^x$
• $g(x) = x^{\left(1 + \frac{1}{x}\right)}$

To find the derivative $\frac{dy}{dx}$, we must compute $\frac{df}{dx}$ and $\frac{dg}{dx}$ separately using logarithmic differentiation [Per the property $\frac{d}{dx}[u^v] = u^v \left( \frac{v}{u} \cdot \frac{du}{dx} + \ln(u) \cdot \frac{dv}{dx} \right)$].

Step 2: Differentiating $f(x) = \left(x + \frac{1}{x}\right)^x$

Taking the natural logarithm on both sides:

$\ln(f) = x \ln\left(x + \frac{1}{x}\right) = x \ln\left(\frac{x^2 + 1}{x}\right) = x [\ln(x^2 + 1) - \ln(x)]$

Differentiating with respect to $x$ using the Product Rule [$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$]:

$\frac{1}{f} \frac{df}{dx} = (x) \cdot \frac{d}{dx}[\ln(x^2 + 1) - \ln(x)] + [\ln(x^2 + 1) - \ln(x)] \cdot \frac{d}{dx}(x)$

$\frac{1}{f} \frac{df}{dx} = x \left( \frac{2x}{x^2 + 1} - \frac{1}{x} \right) + \ln\left(\frac{x^2 + 1}{x}\right)$

$\frac{df}{dx} = f(x) \left[ \frac{2x^2}{x^2 + 1} - 1 + \ln\left(x + \frac{1}{x}\right) \right] = \left(x + \frac{1}{x}\right)^x \left[ \frac{x^2 - 1}{x^2 + 1} + \ln\left(x + \frac{1}{x}\right) \right]$

Step 3: Differentiating $g(x) = x^{\left(1 + \frac{1}{x}\right)}$

Taking the natural logarithm on both sides:

$\ln(g) = \left(1 + \frac{1}{x}\right) \ln(x)$

Differentiating with respect to $x$:

$\frac{1}{g} \frac{dg}{dx} = \left(1 + \frac{1}{x}\right) \cdot \frac{d}{dx}(\ln x) + \ln(x) \cdot \frac{d}{dx}\left(1 + \frac{1}{x}\right)$

$\frac{1}{g} \frac{dg}{dx} = \left(\frac{x + 1}{x}\right) \cdot \frac{1}{x} + \ln(x) \cdot \left(-\frac{1}{x^2}\right)$

$\frac{dg}{dx} = g(x) \left[ \frac{x + 1}{x^2} - \frac{\ln x}{x^2} \right] = x^{\left(1 + \frac{1}{x}\right)} \left[ \frac{x + 1 - \ln x}{x^2} \right]$

Step 4: Combining Results

Since $y = f(x) + g(x)$, then $\frac{dy}{dx} = \frac{df}{dx} + \frac{dg}{dx}$.

Final Solution:

$\frac{dy}{dx} = \left(x + \frac{1}{x}\right)^x \left[ \frac{x^2 - 1}{x^2 + 1} + \ln\left(x + \frac{1}{x}\right) \right] + x^{\left(1 + \frac{1}{x}\right)} \left[ \frac{x + 1 - \ln x}{x^2} \right]$

Step 1: Decomposition of the Function

Let the given function be $y = u + v$, where:

• $u = (x \cos x)^x$
• $v = (x \sin x)^{\frac{1}{x}}$

We are required to find $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$. [Per the Sum Rule of Differentiation: $\frac{d}{dx}(u+v) = \frac{du}{dx} + \frac{dv}{dx}$]

Step 2: Differentiating $u = (x \cos x)^x$

Taking the natural logarithm on both sides: $\ln u = \ln(x \cos x)^x$.
Using the logarithmic property $\ln(a^b) = b \ln a$:
$\ln u = x \ln(x \cos x)$

Differentiating both sides with respect to $x$ using the Product Rule [$\frac{d}{dx}(fg) = f'g + fg'$] and Chain Rule [$\frac{d}{dx}(\ln f(x)) = \frac{1}{f(x)} \cdot f'(x)$]:

$\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}[x] \cdot \ln(x \cos x) + x \cdot \frac{d}{dx}[\ln(x \cos x)]$

$\frac{1}{u} \frac{du}{dx} = 1 \cdot \ln(x \cos x) + x \cdot \left( \frac{1}{x \cos x} \cdot \frac{d}{dx}(x \cos x) \right)$

Using the Product Rule for $\frac{d}{dx}(x \cos x) = \cos x - x \sin x$:

$\frac{1}{u} \frac{du}{dx} = \ln(x \cos x) + \frac{x(\cos x - x \sin x)}{x \cos x}$

$\frac{du}{dx} = (x \cos x)^x \left[ \ln(x \cos x) + \frac{\cos x - x \sin x}{\cos x} \right]$

$\frac{du}{dx} = (x \cos x)^x \left[ \ln(x \cos x) + 1 - x \tan x \right]$

Step 3: Differentiating $v = (x \sin x)^{\frac{1}{x}}$

Taking the natural logarithm: $\ln v = \ln(x \sin x)^{\frac{1}{x}} = \frac{1}{x} \ln(x \sin x)$.
Differentiating both sides with respect to $x$:

$\frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}\left(\frac{1}{x}\right) \ln(x \sin x) + \frac{1}{x} \frac{d}{dx}[\ln(x \sin x)]$

$\frac{1}{v} \frac{dv}{dx} = -\frac{1}{x^2} \ln(x \sin x) + \frac{1}{x} \left( \frac{1}{x \sin x} \cdot \frac{d}{dx}(x \sin x) \right)$

Since $\frac{d}{dx}(x \sin x) = \sin x + x \cos x$:

$\frac{1}{v} \frac{dv}{dx} = -\frac{\ln(x \sin x)}{x^2} + \frac{\sin x + x \cos x}{x^2 \sin x}$

$\frac{dv}{dx} = (x \sin x)^{\frac{1}{x}} \left[ \frac{\sin x + x \cos x - \sin x \ln(x \sin x)}{x^2 \sin x} \right]$

$\frac{dv}{dx} = (x \sin x)^{\frac{1}{x}} \left[ \frac{1 + x \cot x - \ln(x \sin x)}{x^2} \right]$

Step 4: Final Compilation

Summing $\frac{du}{dx}$ and $\frac{dv}{dx}$:

$\frac{dy}{dx} = (x \cos x)^x [1 - x \tan x + \ln(x \cos x)] + (x \sin x)^{\frac{1}{x}} \left[ \frac{1 + x \cot x - \ln(x \sin x)}{x^2} \right]$

Final Solution: $\frac{dy}{dx} = (x \cos x)^x [1 - x \tan x + \ln(x \cos x)] + (x \sin x)^{\frac{1}{x}} \left[ \frac{1 + x \cot x - \ln(x \sin x)}{x^2} \right]$

Step 1: Defining the Function and Sub-problems

Let the given function be $y = (\log x)^x + x^{\log x}$. To differentiate this sum, we define two separate functions, $u$ and $v$, such that $y = u + v$.

$u = (\log x)^x$

$v = x^{\log x}$

The derivative with respect to $x$ is given by $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ [Per the Sum Rule of Differentiation].

Step 2: Differentiating $u = (\log x)^x$

Taking the natural logarithm on both sides: $\log u = \log((\log x)^x)$

Using the power rule of logarithms $\log(a^b) = b \log a$: $\log u = x \cdot \log(\log x)$

Differentiating both sides with respect to $x$ using the Product Rule $(fg)' = f'g + fg'$ and the Chain Rule:

$\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}[x] \cdot \log(\log x) + x \cdot \frac{d}{dx}[\log(\log x)]$

$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log(\log x) + x \cdot \left( \frac{1}{\log x} \cdot \frac{d}{dx}[\log x] \right)$

$\frac{1}{u} \frac{du}{dx} = \log(\log x) + x \cdot \left( \frac{1}{\log x} \cdot \frac{1}{x} \right)$

$\frac{du}{dx} = u \left[ \log(\log x) + \frac{1}{\log x} \right]$

Substituting $u = (\log x)^x$: $\frac{du}{dx} = (\log x)^x \left[ \log(\log x) + \frac{1}{\log x} \right]$

Step 3: Differentiating $v = x^{\log x}$

Taking the natural logarithm on both sides: $\log v = \log(x^{\log x})$

$\log v = \log x \cdot \log x = (\log x)^2$

Differentiating with respect to $x$ using the Chain Rule:

$\frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}[(\log x)^2]$

$\frac{1}{v} \frac{dv}{dx} = 2(\log x) \cdot \frac{d}{dx}[\log x]$

$\frac{1}{v} \frac{dv}{dx} = 2(\log x) \cdot \frac{1}{x}$

$\frac{dv}{dx} = v \cdot \left( \frac{2 \log x}{x} \right)$

Substituting $v = x^{\log x}$: $\frac{dv}{dx} = x^{\log x} \cdot \frac{2 \log x}{x}$

Applying exponent rules $x^a / x^b = x^{a-b}$: $\frac{dv}{dx} = 2 \cdot (\log x) \cdot x^{\log x - 1}$

Step 4: Consolidating the Derivative

Combining the results from Step 2 and Step 3 into $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$:

$\frac{dy}{dx} = (\log x)^x \left[ \log(\log x) + \frac{1}{\log x} \right] + 2 \log x \cdot x^{\log x - 1}$

Final Solution: $\frac{dy}{dx} = (\log x)^x \left[ \log(\log x) + \frac{1}{\log x} \right] + 2 x^{\log x - 1} \log x$

Step 1: Establishing the Product Rule Framework

To differentiate the product of two functions $u(x)$ and $v(x)$, we employ the Product Rule of differentiation, defined as follows [Per Leibniz's Rule for Differentiation]:

$\frac{d}{dx}[u(x) \cdot v(x)] = u(x) \cdot \frac{d}{dx}[v(x)] + v(x) \cdot \frac{d}{dx}[u(x)]$

Let the given function $y$ be defined as:

$y = (x^2 - 5x + 8)(x^3 + 7x + 9)$

Where:

• $u(x) = x^2 - 5x + 8$
• $v(x) = x^3 + 7x + 9$

Step 2: Calculating Individual Derivatives

Applying the Power Rule $\frac{d}{dx}(x^n) = nx^{n-1}$ [By the Power Rule of Calculus]:

$u'(x) = \frac{d}{dx}(x^2 - 5x + 8) = 2x - 5$

$v'(x) = \frac{d}{dx}(x^3 + 7x + 9) = 3x^2 + 7$

Step 3: Substitution into the Product Rule Formula

Substituting $u, v, u',$ and $v'$ into the product rule expression:

$\frac{dy}{dx} = (x^2 - 5x + 8)(3x^2 + 7) + (x^3 + 7x + 9)(2x - 5)$

Step 4: Algebraic Expansion

Expand the terms individually to simplify the expression [By the Distributive Property of Multiplication over Addition]:

Term 1 Expansion: $(x^2 - 5x + 8)(3x^2 + 7) = 3x^4 + 7x^2 - 15x^3 - 35x + 24x^2 + 56 = 3x^4 - 15x^3 + 31x^2 - 35x + 56$

Term 2 Expansion: $(x^3 + 7x + 9)(2x - 5) = 2x^4 - 5x^3 + 14x^2 - 35x + 18x - 45 = 2x^4 - 5x^3 + 14x^2 - 17x - 45$

Step 5: Summation of Terms

Group and combine like terms:

$\frac{dy}{dx} = (3x^4 + 2x^4) + (-15x^3 - 5x^3) + (31x^2 + 14x^2) + (-35x - 17x) + (56 - 45)$

$\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$

Final Solution: \frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11

Step 1: Expansion of the Polynomial Product

Given the function $y = (x^2 - 5x + 8)(x^3 + 7x + 9)$, we apply the distributive property of multiplication over addition [Distributive Law of Algebra]. Each term in the first polynomial is multiplied by each term in the second:

$y = x^2(x^3 + 7x + 9) - 5x(x^3 + 7x + 9) + 8(x^3 + 7x + 9)$

Perform the algebraic expansion:

• $x^2(x^3 + 7x + 9) = x^5 + 7x^3 + 9x^2$
• $-5x(x^3 + 7x + 9) = -5x^4 - 35x^2 - 45x$
• $8(x^3 + 7x + 9) = 8x^3 + 56x + 72$

Step 2: Combining Like Terms

Aggregate the terms by descending order of power [Associative and Commutative Property of Addition]:

$y = x^5 - 5x^4 + (7x^3 + 8x^3) + (9x^2 - 35x^2) + (-45x + 56x) + 72$

$y = x^5 - 5x^4 + 15x^3 - 26x^2 + 11x + 72$

Step 3: Differentiation using Power Rule

To find the derivative $\frac{dy}{dx}$, apply the Sum Rule for differentiation [Sum Rule: $\frac{d}{dx}(u+v) = \frac{du}{dx} + \frac{dv}{dx}$] and the Power Rule [Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$].

$\frac{dy}{dx} = \frac{d}{dx}(x^5) - \frac{d}{dx}(5x^4) + \frac{d}{dx}(15x^3) - \frac{d}{dx}(26x^2) + \frac{d}{dx}(11x) + \frac{d}{dx}(72)$

Step 4: Final Expression Consolidation

Summing the individual derivatives derived in Step 3:

$\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$

Final Solution: $\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$

Step 1: Establishing the Logarithmic Framework

Given the function $y = (x^2 - 5x + 8)(x^3 + 7x + 9)$, we apply the natural logarithm to both sides to simplify the product into a sum of logarithms.

$\ln(y) = \ln((x^2 - 5x + 8)(x^3 + 7x + 9))$

[By the property of logarithms: $\ln(ab) = \ln(a) + \ln(b)$]

$\ln(y) = \ln(x^2 - 5x + 8) + \ln(x^3 + 7x + 9)$

Step 2: Differentiating with respect to $x$

Differentiate both sides with respect to $x$ using the chain rule for the logarithmic terms, $\frac{d}{dx}(\ln(u)) = \frac{1}{u} \cdot \frac{du}{dx}$.

$\frac{d}{dx}(\ln(y)) = \frac{d}{dx}(\ln(x^2 - 5x + 8)) + \frac{d}{dx}(\ln(x^3 + 7x + 9))$

$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{x^2 - 5x + 8} \cdot \frac{d}{dx}(x^2 - 5x + 8) + \frac{1}{x^3 + 7x + 9} \cdot \frac{d}{dx}(x^3 + 7x + 9)$

Step 3: Evaluating Derivatives and Isolating $\frac{dy}{dx}$

Calculate the inner derivatives:

$\frac{d}{dx}(x^2 - 5x + 8) = 2x - 5$

$\frac{d}{dx}(x^3 + 7x + 9) = 3x^2 + 7$

Substitute these into the equation:

$\frac{1}{y} \frac{dy}{dx} = \frac{2x - 5}{x^2 - 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9}$

Isolate $\frac{dy}{dx}$ by multiplying by $y$:

$\frac{dy}{dx} = (x^2 - 5x + 8)(x^3 + 7x + 9) \left[ \frac{2x - 5}{x^2 - 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9} \right]$

Step 4: Final Algebraic Expansion

Multiply $y$ back into the brackets to reach the standard polynomial form:

$\frac{dy}{dx} = (2x - 5)(x^3 + 7x + 9) + (3x^2 + 7)(x^2 - 5x + 8)$

Expanding the terms:

$(2x^4 + 14x^2 + 18x - 5x^3 - 35x - 45) + (3x^4 - 15x^3 + 24x^2 + 7x^2 - 35x + 56)$

Grouping like terms:

$\frac{dy}{dx} = (2x^4 + 3x^4) + (-5x^3 - 15x^3) + (14x^2 + 24x^2 + 7x^2) + (18x - 35x - 35x) + (-45 + 56)$

$\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$

Step 5: Verification of Multi-Method Consistency

Whether using the Product Rule, direct polynomial expansion, or logarithmic differentiation, the fundamental laws of calculus dictate that the derivative of a differentiable function is unique [Uniqueness Theorem of Derivatives].

Final Solution: Yes, all three methods yield the same result: $\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$

Step 1: Decomposition of the Function

Let the given function be denoted by $y$. Since the function is a difference of two variable-dependent terms, we define:

$y = u - v$

Where $u = x^x$ and $v = 2^{\sin x}$.

The derivative of the function with respect to $x$ is given by the linearity of the derivative operator: [Per the Sum/Difference Rule of Differentiation]

$\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$

Step 2: Differentiating $u = x^x$

To differentiate a function where both the base and the exponent are variables, we utilize logarithmic differentiation [Per the property $\ln(a^b) = b \ln a$]:

$\ln u = \ln(x^x)$

$\ln u = x \ln x$

Differentiating both sides with respect to $x$ using the Product Rule [$\frac{d}{dx}(fg) = f'g + fg'$]:

$\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) \cdot \ln x + x \cdot \frac{d}{dx}(\ln x)$

$\frac{1}{u} \frac{du}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x}$

$\frac{1}{u} \frac{du}{dx} = \ln x + 1$

$\frac{du}{dx} = u(1 + \ln x) = x^x(1 + \ln x)$

Step 3: Differentiating $v = 2^{\sin x}$

We use the exponential differentiation formula $\frac{d}{dx}(a^f(x)) = a^f(x) \ln a \cdot f'(x)$ [Per the Chain Rule applied to $a^u$]:

$v = 2^{\sin x}$

$\frac{dv}{dx} = 2^{\sin x} \cdot \ln 2 \cdot \frac{d}{dx}(\sin x)$

Since $\frac{d}{dx}(\sin x) = \cos x$:

$\frac{dv}{dx} = 2^{\sin x} \cdot \ln 2 \cdot \cos x$

Step 4: Synthesis of the Final Result

Substituting the results from Step 2 and Step 3 into the expression derived in Step 1:

$\frac{dy}{dx} = x^x(1 + \ln x) - 2^{\sin x} \cdot \cos x \cdot \ln 2$

Final Solution: $\frac{dy}{dx} = x^x(1 + \ln x) - (\cos x) 2^{\sin x} \ln 2$

Method 1: Repeated Application of the Product Rule

Given the product of three functions $f(x) = u \cdot v \cdot w$, we treat $(u \cdot v)$ as a single entity and apply the product rule $[\frac{d}{dx}(f \cdot g) = f'g + fg']$:

Step 1: Let $y = (u \cdot v) \cdot w$. Applying the product rule:

$\frac{dy}{dx} = \frac{d}{dx}(u \cdot v) \cdot w + (u \cdot v) \cdot \frac{d}{dx}(w)$ [By Product Rule]

Step 2: Expand the derivative of the product $(u \cdot v)$:

$\frac{dy}{dx} = \left( \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx} \right) \cdot w + u \cdot v \cdot \frac{dw}{dx}$ [By Product Rule applied to $(u \cdot v)$]

Step 3: Distribute $w$ into the parenthesis:

$\frac{dy}{dx} = \frac{du}{dx} \cdot v \cdot w + u \cdot \frac{dv}{dx} \cdot w + u \cdot v \cdot \frac{dw}{dx}$

Method 2: Logarithmic Differentiation

Step 1: Take the natural logarithm on both sides of the equation $y = u \cdot v \cdot w$:

$\ln(y) = \ln(u \cdot v \cdot w)$

Step 2: Apply the logarithmic property $\ln(abc) = \ln(a) + \ln(b) + \ln(c)$:

$\ln(y) = \ln(u) + \ln(v) + \ln(w)$

Step 3: Differentiate both sides with respect to $x$:

$\frac{d}{dx}[\ln(y)] = \frac{d}{dx}[\ln(u)] + \frac{d}{dx}[\ln(v)] + \frac{d}{dx}[\ln(w)]$

Step 4: Apply the chain rule $\frac{d}{dx}\ln(f) = \frac{1}{f} \cdot f'$:

$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} + \frac{1}{v} \cdot \frac{dv}{dx} + \frac{1}{w} \cdot \frac{dw}{dx}$

Step 5: Multiply both sides by $y$:

$\frac{dy}{dx} = y \left( \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx} \right)$

Step 6: Substitute $y = u \cdot v \cdot w$ back into the equation:

$\frac{dy}{dx} = (u \cdot v \cdot w) \left( \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx} \right)$

$\frac{dy}{dx} = (u \cdot v \cdot w \cdot \frac{1}{u} \frac{du}{dx}) + (u \cdot v \cdot w \cdot \frac{1}{v} \frac{dv}{dx}) + (u \cdot v \cdot w \cdot \frac{1}{w} \frac{dw}{dx})$

$\frac{dy}{dx} = \frac{du}{dx} \cdot v \cdot w + u \cdot \frac{dv}{dx} \cdot w + u \cdot v \cdot \frac{dw}{dx}$

Visual Representation of Derivative Summation

Final Solution: $\frac{d}{dx} (u \cdot v \cdot w) = \frac{du}{dx} v \cdot w + u \cdot \frac{dv}{dx} \cdot w + u \cdot v \cdot \frac{dw}{dx}$

Step 1: Assigning Variables and Logarithmic Transformation

Let the given function be represented as:

$y = (\log x)^{\cos x}$

To differentiate a function where both the base and the exponent are variables, we apply the natural logarithm to both sides. [By the property $\log(a^b) = b \cdot \log a$]:

$\log y = \log((\log x)^{\cos x})$

$\log y = \cos x \cdot \log(\log x)$

Step 2: Differentiation with respect to $x$

Differentiate both sides with respect to $x$ using the Chain Rule and the Product Rule [Per the Product Rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$]:

$\frac{d}{dx}(\log y) = \frac{d}{dx}(\cos x \cdot \log(\log x))$

$\frac{1}{y} \frac{dy}{dx} = \cos x \cdot \frac{d}{dx}(\log(\log x)) + \log(\log x) \cdot \frac{d}{dx}(\cos x)$

Step 3: Evaluating Individual Derivatives

Applying the Chain Rule for $\frac{d}{dx}(\log(\log x))$:

$\frac{d}{dx}(\log(\log x)) = \frac{1}{\log x} \cdot \frac{d}{dx}(\log x) = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}$

For the second term:

$\frac{d}{dx}(\cos x) = -\sin x$

Step 4: Algebraic Substitution and Solving for $\frac{dy}{dx}$

Substitute the evaluated derivatives back into the equation:

$\frac{1}{y} \frac{dy}{dx} = \cos x \left( \frac{1}{x \log x} \right) + \log(\log x) (-\sin x)$

$\frac{dy}{dx} = y \left( \frac{\cos x}{x \log x} - \sin x \log(\log x) \right)$

Step 5: Final Substitution of $y$

Substitute $y = (\log x)^{\cos x}$ to express the derivative in terms of $x$:

$\frac{dy}{dx} = (\log x)^{\cos x} \left[ \frac{\cos x}{x \log x} - \sin x \log(\log x) \right]$

Final Solution: $\frac{dy}{dx} = (\log x)^{\cos x} \left[ \frac{\cos x}{x \log x} - \sin x \log(\log x) \right]$

Step 1: Application of Logarithmic Differentiation

Given the implicit function: $y^x = x^y$

To differentiate variables raised to the power of variables, we apply the natural logarithm ($\ln$) to both sides of the equation [Per the property $\ln(a^b) = b \ln a$]:

$\ln(y^x) = \ln(x^y)$

$x \ln y = y \ln x$

Step 2: Differentiating with Respect to $x$

We differentiate both sides with respect to $x$ using the Product Rule, which states $\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$ [By the Product Rule of Differentiation]:

$\frac{d}{dx}(x \ln y) = \frac{d}{dx}(y \ln x)$

$x \cdot \frac{d}{dx}(\ln y) + \ln y \cdot \frac{d}{dx}(x) = y \cdot \frac{d}{dx}(\ln x) + \ln x \cdot \frac{d}{dx}(y)$

Step 3: Executing Chain Rule and Implicit Differentiation

Apply the Chain Rule $\frac{d}{dx}[\ln(f(x))] = \frac{1}{f(x)} \cdot f'(x)$ [Per Chain Rule]:

$x \left( \frac{1}{y} \cdot \frac{dy}{dx} \right) + \ln y (1) = y \left( \frac{1}{x} \right) + \ln x \left( \frac{dy}{dx} \right)$

$\frac{x}{y} \frac{dy}{dx} + \ln y = \frac{y}{x} + \ln x \frac{dy}{dx}$

Step 4: Isolating the $\frac{dy}{dx}$ Term

Group all terms containing $\frac{dy}{dx}$ on the left-hand side and constants on the right-hand side:

$\frac{x}{y} \frac{dy}{dx} - \ln x \frac{dy}{dx} = \frac{y}{x} - \ln y$

$\frac{dy}{dx} \left( \frac{x}{y} - \ln x \right) = \frac{y - x \ln y}{x}$

Step 5: Solving for the Final Derivative

Simplify the bracketed term to find the common denominator $y$:

$\frac{dy}{dx} \left( \frac{x - y \ln x}{y} \right) = \frac{y - x \ln y}{x}$

Multiply both sides by $\left( \frac{y}{x - y \ln x} \right)$:

$\frac{dy}{dx} = \frac{y(y - x \ln y)}{x(x - y \ln x)}$

Final Solution: $\frac{dy}{dx} = \frac{y(y - x \ln y)}{x(x - y \ln x)}$

Step 1: Logarithmic Transformation of the Equation

Given the equation: $xy = e^{(x - y)}$

To isolate the variables, we apply the natural logarithm ($\ln$) to both sides of the equation [Per the property $\ln(a^b) = b \ln a$ and $\ln(e^x) = x$]:

$\ln(xy) = \ln(e^{x - y})$

Using the logarithmic property $\ln(ab) = \ln(a) + \ln(b)$, we expand the left side:

$\ln(x) + \ln(y) = x - y$

Step 2: Implicit Differentiation with respect to $x$

Differentiate both sides of the equation with respect to $x$ using the chain rule and the derivative rules for logarithmic functions [Per $\frac{d}{dx}(\ln x) = \frac{1}{x}$ and $\frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx}$]:

$\frac{d}{dx}(\ln x) + \frac{d}{dx}(\ln y) = \frac{d}{dx}(x) - \frac{d}{dx}(y)$

$\frac{1}{x} + \frac{1}{y} \cdot \frac{dy}{dx} = 1 - \frac{dy}{dx}$

Step 3: Algebraic Isolation of $\frac{dy}{dx}$

Gather all terms containing $\frac{dy}{dx}$ on the left-hand side of the equation:

$\frac{1}{y} \cdot \frac{dy}{dx} + \frac{dy}{dx} = 1 - \frac{1}{x}$

Factor out $\frac{dy}{dx}$ [By the Distributive Property]:

$\frac{dy}{dx} \left( \frac{1}{y} + 1 \right) = \frac{x - 1}{x}$

Simplify the term in the parentheses:

$\frac{dy}{dx} \left( \frac{1 + y}{y} \right) = \frac{x - 1}{x}$

Step 4: Final Expression Construction

Solve for $\frac{dy}{dx}$ by multiplying both sides by the reciprocal of the term in parentheses:

$\frac{dy}{dx} = \left( \frac{x - 1}{x} \right) \cdot \left( \frac{y}{1 + y} \right)$

$\frac{dy}{dx} = \frac{y(x - 1)}{x(1 + y)}$

Final Solution: $\frac{dy}{dx} = \frac{y(x - 1)}{x(y + 1)}$

Step 1: Decomposition of the Function

To differentiate the expression $y = (\sin x)^x + \sin^{-1}(\sqrt{x})$, we define two auxiliary functions, $u$ and $v$, such that:

$y = u + v$

where $u = (\sin x)^x$ and $v = \sin^{-1}(\sqrt{x})$

[By the Linearity of the Derivative Operator: $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$]

Step 2: Differentiating $u = (\sin x)^x$

Taking the natural logarithm on both sides:

$\ln(u) = \ln((\sin x)^x)$

[By Logarithmic Property: $\ln(a^b) = b \cdot \ln(a)$]

$\ln(u) = x \cdot \ln(\sin x)$

Differentiating both sides with respect to $x$ using the Product Rule and Chain Rule:

$\frac{1}{u} \cdot \frac{du}{dx} = \frac{d}{dx}[x] \cdot \ln(\sin x) + x \cdot \frac{d}{dx}[\ln(\sin x)]$

[By Product Rule: $(fg)' = f'g + fg'$]

$\frac{1}{u} \cdot \frac{du}{dx} = 1 \cdot \ln(\sin x) + x \cdot \left( \frac{1}{\sin x} \cdot \cos x \right)$

[By Chain Rule: $\frac{d}{dx}(\ln(f(x))) = \frac{1}{f(x)}f'(x)$]

$\frac{du}{dx} = u \cdot [\ln(\sin x) + x \cot x]$

$\frac{du}{dx} = (\sin x)^x \cdot [\ln(\sin x) + x \cot x]$

Step 3: Differentiating $v = \sin^{-1}(\sqrt{x})$

Applying the standard derivative formula for inverse trigonometric functions:

[Per the derivative identity: $\frac{d}{dx}(\sin^{-1} u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$]

$\frac{dv}{dx} = \frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x})$

$\frac{dv}{dx} = \frac{1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}}$

[Using Power Rule: $\frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$]

$\frac{dv}{dx} = \frac{1}{2\sqrt{x(1-x)}} = \frac{1}{2\sqrt{x-x^2}}$

Step 4: Combining the Results

Summing the individual derivatives to obtain the total derivative $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$

Final Solution: $\frac{dy}{dx} = (\sin x)^x [\ln(\sin x) + x \cot x] + \frac{1}{2\sqrt{x-x^2}}$

Step 1: Taking the Logarithm of Both Sides

Given the equation: $(\cos x)^y = (\cos y)^x$

To differentiate functions where variables appear in both the base and the exponent, we apply the natural logarithm ($\ln$) to both sides [By the property of logarithmic functions]:

$\ln((\cos x)^y) = \ln((\cos y)^x)$

Using the logarithmic power rule $\ln(a^b) = b \ln(a)$, we simplify the expression:

$y \ln(\cos x) = x \ln(\cos y)$

Step 2: Implicit Differentiation

We differentiate both sides with respect to $x$, applying the product rule $\frac{d}{dx}[uv] = u \frac{dv}{dx} + v \frac{du}{dx}$ and the chain rule [Per the Chain Rule for differentiation]:

$\frac{d}{dx}[y \ln(\cos x)] = \frac{d}{dx}[x \ln(\cos y)]$

$y \cdot \frac{d}{dx}[\ln(\cos x)] + \ln(\cos x) \cdot \frac{dy}{dx} = x \cdot \frac{d}{dx}[\ln(\cos y)] + \ln(\cos y) \cdot \frac{d}{dx}[x]$

Step 3: Evaluating Derivatives

Applying the derivative of $\ln(u)$ which is $\frac{1}{u} \cdot \frac{du}{dx}$:

$y \cdot \left( \frac{1}{\cos x} \cdot (-\sin x) \right) + \ln(\cos x) \frac{dy}{dx} = x \cdot \left( \frac{1}{\cos y} \cdot (-\sin y) \frac{dy}{dx} \right) + \ln(\cos y) \cdot 1$

Simplifying the trigonometric terms ($\frac{\sin \theta}{\cos \theta} = \tan \theta$):

$-y \tan x + \ln(\cos x) \frac{dy}{dx} = -x \tan y \frac{dy}{dx} + \ln(\cos y)$

Step 4: Isolating the $\frac{dy}{dx}$ Terms

Group all terms containing $\frac{dy}{dx}$ on the left-hand side and the remaining terms on the right-hand side:

$\ln(\cos x) \frac{dy}{dx} + x \tan y \frac{dy}{dx} = \ln(\cos y) + y \tan x$

Factor out $\frac{dy}{dx}$:

$\frac{dy}{dx} [\ln(\cos x) + x \tan y] = \ln(\cos y) + y \tan x$

Step 5: Final Algebraic Solution

Divide both sides by the bracketed coefficient to solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{\ln(\cos y) + y \tan x}{\ln(\cos x) + x \tan y}$

Final Solution: $\frac{dy}{dx} = \frac{\ln(\cos y) + y \tan x}{\ln(\cos x) + x \tan y}$

Step 1: Establishing the Equation and Applying Logarithms

Let the given function be represented as $y$: $y = \cos x \cdot \cos 2x \cdot \cos 3x$

Since this function is a product of multiple trigonometric functions, logarithmic differentiation is the most efficient method to simplify the expression [Property: $\ln(abc) = \ln a + \ln b + \ln c$].

Taking the natural logarithm ($\ln$) on both sides: $\ln y = \ln(\cos x \cdot \cos 2x \cdot \cos 3x)$

Applying the product rule for logarithms: $\ln y = \ln(\cos x) + \ln(\cos 2x) + \ln(\cos 3x)$

Step 2: Differentiation with Respect to $x$

Differentiate both sides with respect to $x$ using the Chain Rule [Per the Chain Rule: $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$]:

$\frac{d}{dx}(\ln y) = \frac{d}{dx}(\ln(\cos x)) + \frac{d}{dx}(\ln(\cos 2x)) + \frac{d}{dx}(\ln(\cos 3x))$

Applying the derivative of $\ln(u)$ which is $\frac{1}{u} \cdot \frac{du}{dx}$:

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{\cos x} \cdot (-\sin x) + \frac{1}{\cos 2x} \cdot (-\sin 2x \cdot 2) + \frac{1}{\cos 3x} \cdot (-\sin 3x \cdot 3)$

Step 3: Algebraic Simplification

Utilizing the trigonometric identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

$\frac{1}{y} \frac{dy}{dx} = -\tan x - 2\tan 2x - 3\tan 3x$

Isolating $\frac{dy}{dx}$ by multiplying both sides by $y$:

$\frac{dy}{dx} = -y (\tan x + 2\tan 2x + 3\tan 3x)$

Step 4: Substitution of the Original Function

Substitute $y = \cos x \cdot \cos 2x \cdot \cos 3x$ back into the equation:

$\frac{dy}{dx} = -(\cos x \cdot \cos 2x \cdot \cos 3x)(\tan x + 2\tan 2x + 3\tan 3x)$

Final Solution: $\frac{dy}{dx} = -(\cos x \cdot \cos 2x \cdot \cos 3x)(\tan x + 2\tan 2x + 3\tan 3x)$