Step 1: Application of the Product Rule

Given the function $f(x) = \cos(x^3) \cdot \sin^2(x^5)$, we designate the components as:

• $u(x) = \cos(x^3)$
• $v(x) = \sin^2(x^5) = [\sin(x^5)]^2$

To differentiate the product, we utilize the Product Rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$ [By the Leibniz Product Rule].

Step 2: Differentiation of Individual Components

A. Differentiating $u(x) = \cos(x^3)$ using the Chain Rule:

Let $t = x^3$, then $\frac{dt}{dx} = 3x^2$.

$u'(x) = \frac{d}{dt}(\cos t) \cdot \frac{dt}{dx} = -\sin(x^3) \cdot 3x^2 = -3x^2 \sin(x^3)$ [By the Chain Rule].

B. Differentiating $v(x) = [\sin(x^5)]^2$ using the Chain Rule:

Let $w = \sin(x^5)$, then $v(x) = w^2$.

$v'(x) = \frac{d}{dw}(w^2) \cdot \frac{d}{dx}(\sin(x^5)) = 2w \cdot \cos(x^5) \cdot \frac{d}{dx}(x^5)$ [By the Chain Rule].

$v'(x) = 2 \sin(x^5) \cdot \cos(x^5) \cdot 5x^4 = 10x^4 \sin(x^5) \cos(x^5)$ [Using the algebraic derivative rule $\frac{d}{dx}x^n = nx^{n-1}$].

Step 3: Synthesis of Derivatives

Substituting the results from Step 2 into the Product Rule formula:

$\frac{dy}{dx} = [-3x^2 \sin(x^3)] \cdot [\sin^2(x^5)] + [\cos(x^3)] \cdot [10x^4 \sin(x^5) \cos(x^5)]$

Step 4: Algebraic Simplification

The derivative expression is:

$\frac{dy}{dx} = -3x^2 \sin(x^3) \sin^2(x^5) + 10x^4 \cos(x^3) \sin(x^5) \cos(x^5)$

We may factor out common terms, such as $x^2 \sin(x^5)$, to reach a cleaner analytical form:

$\frac{dy}{dx} = x^2 \sin(x^5) \left[ 10x^2 \cos(x^3) \cos(x^5) - 3 \sin(x^3) \sin(x^5) \right]$

Final Solution: $\frac{dy}{dx} = x^2 \sin(x^5) [10x^2 \cos(x^3) \cos(x^5) - 3 \sin(x^3) \sin(x^5)]$

Step 1: Formal Setup of the Derivative

Given the function: $f(x) = 2\sqrt{\cot(x^2)}$. We are required to find the derivative with respect to $x$, denoted as $\frac{d}{dx}[2\sqrt{\cot(x^2)}]$.

By the Constant Multiple Rule: $\frac{d}{dx}[c \cdot f(x)] = c \cdot \frac{d}{dx}[f(x)]$. Thus:

$$\frac{dy}{dx} = 2 \cdot \frac{d}{dx} \left[ (\cot(x^2))^{1/2} \right]$$

Step 2: Application of the Chain Rule

Let $u = \cot(x^2)$. Then the expression becomes $2\sqrt{u}$. By the Chain Rule [Per the Chain Rule for composite functions $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$]:

$$\frac{dy}{dx} = 2 \cdot \left( \frac{1}{2\sqrt{\cot(x^2)}} \right) \cdot \frac{d}{dx}[\cot(x^2)]$$

Step 3: Differentiation of Inner Functions

We apply the Chain Rule again to find $\frac{d}{dx}[\cot(x^2)]$. Knowing that $\frac{d}{d\theta}[\cot \theta] = -\csc^2 \theta$ [Standard Trigonometric Derivative]:

$$\frac{d}{dx}[\cot(x^2)] = -\csc^2(x^2) \cdot \frac{d}{dx}[x^2]$$

Using the Power Rule $\frac{d}{dx}[x^n] = nx^{n-1}$:

$$\frac{d}{dx}[x^2] = 2x$$

Therefore, $\frac{d}{dx}[\cot(x^2)] = -2x \csc^2(x^2)$.

Step 4: Algebraic Consolidation

Substituting the results back into the expression from Step 2:

$$\frac{dy}{dx} = 2 \cdot \left( \frac{1}{2\sqrt{\cot(x^2)}} \right) \cdot \left( -2x \csc^2(x^2) \right)$$

The constant $2$ in the numerator and the $2$ in the denominator cancel out:

$$\frac{dy}{dx} = \frac{-2x \csc^2(x^2)}{\sqrt{\cot(x^2)}}$$

Step 5: Trigonometric Simplification (Optional)

Using the identity $\cot \theta = \frac{\cos \theta}{\sin \theta}$ and $\csc \theta = \frac{1}{\sin \theta}$:

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

While the expression above is expanded, the standard differential form is generally accepted as follows:

Final Solution: $\frac{dy}{dx} = \frac{-2x \csc^2(x^2)}{\sqrt{\cot(x^2)}}$

Step 1: Formal Representation of the Function

Let the given function be represented as $y = f(x)$. Thus, we define:

$y = \sin(ax + b)$

[Where $a$ and $b$ are constants and $x$ is the independent variable].

Step 2: Identification of the Differentiation Rule

To differentiate this function, we must apply the Chain Rule [Calculus Theorem: If $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$].

Here, the composite function consists of an outer function $f(u) = \sin(u)$ and an inner function $u = g(x) = ax + b$.

Step 3: Application of Derivatives

Differentiating both sides with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx} [\sin(ax + b)]$

Applying the standard derivative formula $\frac{d}{dx}(\sin u) = \cos u \cdot \frac{du}{dx}$:

$\frac{dy}{dx} = \cos(ax + b) \cdot \frac{d}{dx}(ax + b)$

Step 4: Evaluating the Internal Derivative

Now, compute the derivative of the inner function $(ax + b)$ using the Sum Rule and Constant Multiple Rule of derivatives:

$\frac{d}{dx}(ax + b) = \frac{d}{dx}(ax) + \frac{d}{dx}(b)$

Since $a$ is a constant, $\frac{d}{dx}(ax) = a \cdot \frac{d}{dx}(x) = a(1) = a$.

Since $b$ is a constant, $\frac{d}{dx}(b) = 0$ [Derivative of a constant is zero].

Therefore, $\frac{d}{dx}(ax + b) = a$.

Step 5: Final Substitution and Simplification

Substitute the result from Step 4 back into the expression from Step 3:

$\frac{dy}{dx} = \cos(ax + b) \cdot (a)$

$\frac{dy}{dx} = a \cos(ax + b)$

Final Solution: $\frac{dy}{dx} = a \cos(ax + b)$

Step 1: Identifying the Functional Structure

The given function is a composite function of the form $f(g(x))$, where $f(u) = \sin(u)$ and $g(x) = x^2 + 5$. To differentiate such a function, we must employ the Chain Rule [Per the Chain Rule for differentiation, if $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$].

Step 2: Assigning Intermediate Variables

Let $y = \sin(x^2 + 5)$.
Let $u = x^2 + 5$.
Then the function becomes $y = \sin(u)$.

Step 3: Calculating Partial Derivatives

First, differentiate $y$ with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(\sin u) = \cos u$ [Per the standard derivative formula $\frac{d}{dx}(\sin x) = \cos x$].

Second, differentiate $u$ with respect to $x$:

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

Using the Sum Rule and Power Rule [Per $\frac{d}{dx}(x^n) = nx^{n-1}$ and $\frac{d}{dx}(c) = 0$]:

$\frac{du}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}(5) = 2x + 0 = 2x$.

Step 4: Applying the Chain Rule

Combine the derivatives using the relationship $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$:

$\frac{dy}{dx} = \cos(u) \cdot (2x)$

Step 5: Substitution and Final Simplification

Substitute $u = x^2 + 5$ back into the expression:

$\frac{dy}{dx} = \cos(x^2 + 5) \cdot 2x$

Rearranging for standard algebraic notation:

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

Final Solution: $\frac{dy}{dx} = 2x \cos(x^2 + 5)$

Step 1: Formal Definition of the Function

Let the given function be denoted as $y = f(x)$. We define the function as follows:

$y = \cos(\sin x)$

Step 2: Applying the Chain Rule

To differentiate a composite function of the form $f(g(x))$, we employ the Chain Rule [Per the Chain Rule for differentiation: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$].

Let $u = \sin x$. Then, the function becomes $y = \cos u$.

We are required to find $\frac{dy}{dx}$ using the relation:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Step 3: Calculating Individual Derivatives

First, differentiate $y$ with respect to $u$:

$y = \cos u \implies \frac{dy}{du} = -\sin u$ [Using the standard derivative formula $\frac{d}{dx}(\cos x) = -\sin x$]

Second, differentiate $u$ with respect to $x$:

$u = \sin x \implies \frac{du}{dx} = \cos x$ [Using the standard derivative formula $\frac{d}{dx}(\sin x) = \cos x$]

Step 4: Substitution and Simplification

Substitute the derivatives found in Step 3 into the Chain Rule formula defined in Step 2:

$\frac{dy}{dx} = (-\sin u) \cdot (\cos x)$

Now, substitute back $u = \sin x$ to express the final derivative in terms of the independent variable $x$:

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

Visual Representation of Function Composition

Final Solution: $\frac{dy}{dx} = -\cos x \sin(\sin x)$

Given Definition & Initial Setup

The function $f(x) = |x - 1|$ is defined for all $x \in \mathbb{R}$. By the definition of the absolute value function:

$f(x) = \begin{cases} -(x - 1), & \text{if } x < 1 \\ x - 1, & \text{if } x \geq 1 \end{cases}$

For a function $f(x)$ to be differentiable at a point $x = c$, the Left Hand Derivative (LHD) and Right Hand Derivative (RHD) must exist and be equal:
$LHD = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h} = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h} = RHD$

Step 1: Calculating the Left Hand Derivative (LHD) at $x = 1$

Using the limit definition where $h > 0$ and $h \to 0$:

$LHD = \lim_{h \to 0^+} \frac{f(1 - h) - f(1)}{-h}$ [By definition of derivative at $x=c$ with $x = c-h$]

Since $1 - h < 1$, we use the definition $f(x) = -(x - 1)$:

$LHD = \lim_{h \to 0^+} \frac{[-( (1 - h) - 1 )] - |1 - 1|}{-h}$

$LHD = \lim_{h \to 0^+} \frac{-(-h) - 0}{-h} = \lim_{h \to 0^+} \frac{h}{-h} = -1$

Step 2: Calculating the Right Hand Derivative (RHD) at $x = 1$

Using the limit definition where $h > 0$ and $h \to 0$:

$RHD = \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h}$

Since $1 + h > 1$, we use the definition $f(x) = x - 1$:

$RHD = \lim_{h \to 0^+} \frac{( (1 + h) - 1 ) - (1 - 1)}{h}$

$RHD = \lim_{h \to 0^+} \frac{h - 0}{h} = \lim_{h \to 0^+} (1) = 1$

Step 3: Analytical Comparison and Conclusion

We observe that $LHD = -1$ and $RHD = 1$.

Since $LHD \neq RHD$, the limit $\lim_{x \to 1} \frac{f(x) - f(1)}{x - 1}$ does not exist [By the definition of Differentiability at a Point].

Visual Representation of the Non-Differentiable Point

Final Solution: Since the Left Hand Derivative (-1) is not equal to the Right Hand Derivative (1), the function $f(x) = |x - 1|$ is not differentiable at $x = 1$.

Given Variables & Initial Setup

The greatest integer function is defined as $f(x) = [x]$, which represents the greatest integer less than or equal to $x$. We are analyzing the differentiability of $f(x)$ at $x = 1$ and $x = 2$ within the open interval $(0, 3)$.

By definition, a function $f(x)$ is differentiable at a point $x = c$ if and only if the Left-Hand Derivative (LHD) and the Right-Hand Derivative (RHD) exist and are equal:
$\lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h} = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h}$

Step 1: Analyzing Differentiability at $x = 1$

To test differentiability at $x = 1$, we calculate the LHD and RHD where $h > 0$ and $h$ is a very small positive quantity.

• LHD at $x = 1$:
  $LHD = \lim_{h \to 0^+} \frac{f(1-h) - f(1)}{-h}$
  Since $1-h$ is slightly less than 1, $[1-h] = 0$. Since $[1] = 1$:
  $LHD = \lim_{h \to 0^+} \frac{0 - 1}{-h} = \lim_{h \to 0^+} \frac{-1}{-h} = \infty$ [The limit does not exist as a finite value].
• RHD at $x = 1$:
  $RHD = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h}$
  Since $1+h$ is slightly greater than 1, $[1+h] = 1$.
  $RHD = \lim_{h \to 0^+} \frac{1 - 1}{h} = 0$.

Since $LHD \neq RHD$ (and the LHD is infinite), $f(x)$ is not differentiable at $x = 1$.

Step 2: Analyzing Differentiability at $x = 2$

• LHD at $x = 2$:
  $LHD = \lim_{h \to 0^+} \frac{f(2-h) - f(2)}{-h}$
  Since $2-h$ is slightly less than 2, $[2-h] = 1$. Since $[2] = 2$:
  $LHD = \lim_{h \to 0^+} \frac{1 - 2}{-h} = \lim_{h \to 0^+} \frac{-1}{-h} = \infty$.
• RHD at $x = 2$:
  $RHD = \lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h}$
  Since $2+h$ is slightly greater than 2, $[2+h] = 2$.
  $RHD = \lim_{h \to 0^+} \frac{2 - 2}{h} = 0$.

Since $LHD \neq RHD$, $f(x)$ is not differentiable at $x = 2$.

Visual Representation of the Greatest Integer Function

Theoretical Justification

Per the property of jump discontinuities in step functions, the function $f(x) = [x]$ possesses a jump discontinuity at every integer $n \in \mathbb{Z}$. Since differentiability requires continuity at the point of interest [By the Differentiability-Continuity Theorem: if a function is differentiable at $c$, it must be continuous at $c$], and the greatest integer function is discontinuous at integers, it is inherently non-differentiable at $x=1$ and $x=2$.

Final Solution: The function $f(x) = [x]$ is not differentiable at $x = 1$ and $x = 2$ because the Left-Hand Derivative and Right-Hand Derivative do not exist as equal finite values at these points, violating the fundamental condition for differentiability.

Step 1: Identifying the Functional Composition

To differentiate the function $y = \cos(\sqrt{x})$, we must identify this as a composite function of the form $f(g(x))$, where:

• $f(u) = \cos(u)$, where $u = \sqrt{x}$
• $g(x) = \sqrt{x} = x^{1/2}$

[By the Chain Rule of differentiation, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$]

Step 2: Applying the Chain Rule

We perform the differentiation in two distinct stages:

1. Differentiate the outer function, $\cos(u)$, with respect to $u$:

$\frac{d}{du}(\cos(u)) = -\sin(u)$ [Derivative of trigonometric function]

2. Differentiate the inner function, $u = x^{1/2}$, with respect to $x$:

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

Step 3: Synthesis and Simplification

Substitute the components back into the Chain Rule formula:

$\frac{dy}{dx} = -\sin(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x})$

$\frac{dy}{dx} = -\sin(\sqrt{x}) \cdot \left( \frac{1}{2\sqrt{x}} \right)$

Analytical Logic Table

Final Solution: $\frac{dy}{dx} = -\frac{\sin(\sqrt{x})}{2\sqrt{x}}$

Step 1: Identifying the Mathematical Structure

The given function is of the form $f(x) = \frac{u(x)}{v(x)}$, where:

• $u(x) = \sin(ax+b)$
• $v(x) = \cos(cx+d)$

To differentiate this, we apply the Quotient Rule, defined as [Per the Leibniz Quotient Rule]:

$\frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right] = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Step 2: Calculating Individual Derivatives

We apply the Chain Rule [Per the Chain Rule of Differentiation]:

1. For $u(x) = \sin(ax+b)$:

2. For $v(x) = \cos(cx+d)$:

Step 3: Substituting into the Quotient Rule Formula

Substituting the values derived in Step 2 into the Quotient Rule formula:

$\frac{dy}{dx} = \frac{[a \cos(ax+b)] \cdot [\cos(cx+d)] - [\sin(ax+b)] \cdot [-c \sin(cx+d)]}{[\cos(cx+d)]^2}$

Step 4: Algebraic Simplification

Simplify the signs and express the result:

$\frac{dy}{dx} = \frac{a \cos(ax+b) \cos(cx+d) + c \sin(ax+b) \sin(cx+d)}{\cos^2(cx+d)}$

We can further simplify by splitting the fraction:

$\frac{dy}{dx} = \frac{a \cos(ax+b) \cos(cx+d)}{\cos^2(cx+d)} + \frac{c \sin(ax+b) \sin(cx+d)}{\cos^2(cx+d)}$

$\frac{dy}{dx} = a \cos(ax+b) \sec(cx+d) + c \sin(ax+b) \sin(cx+d) \sec^2(cx+d)$

Visual Representation of the Slope Function

Final Solution: $\frac{dy}{dx} = \frac{a \cos(ax+b) \cos(cx+d) + c \sin(ax+b) \sin(cx+d)}{\cos^2(cx+d)}$

Given Function

Let the function be represented by $y$: $y = \sec(\tan(\sqrt{x}))$

Step 1: Applying the Chain Rule

To differentiate $y$ with respect to $x$, we employ the Chain Rule of differentiation. The Chain Rule states that for a composite function $f(g(h(x)))$, the derivative is $f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$. [Per the Chain Rule of Differentiation]

Let $u = \tan(\sqrt{x})$ and $v = \sqrt{x}$. Then $y = \sec(u)$, $u = \tan(v)$, and $v = x^{1/2}$.

Step 2: Differentiating Outer Functions

The derivative of $\sec(\theta)$ is $\sec(\theta)\tan(\theta)$. $\frac{dy}{du} = \sec(\tan(\sqrt{x})) \cdot \tan(\tan(\sqrt{x}))$

Step 3: Differentiating Intermediate Functions

The derivative of $\tan(\theta)$ is $\sec^2(\theta)$. $\frac{du}{dv} = \sec^2(\sqrt{x})$

Step 4: Differentiating Inner Function

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$: $\frac{dv}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

Step 5: Synthesizing the Derivative

Combining the derivatives according to the Chain Rule formula $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$:

$\frac{dy}{dx} = \sec(\tan(\sqrt{x})) \cdot \tan(\tan(\sqrt{x})) \cdot \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$

Visual Representation of the Chain Structure

Final Simplification

Rearranging the terms for standard notation:

$\frac{dy}{dx} = \frac{\sec(\tan(\sqrt{x})) \cdot \tan(\tan(\sqrt{x})) \cdot \sec^2(\sqrt{x})}{2\sqrt{x}}$

Final Solution: \frac{d}{dx}[\sec(\tan(\sqrt{x}))] = \frac{\sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x})) \sec^2(\sqrt{x})}{2\sqrt{x}}