Step 1: Identifying the Mathematical Structure

To differentiate the function $f(x) = \dfrac{e^x}{\sin x}$ with respect to $x$, we identify the form as a quotient of two differentiable functions, $u(x) = e^x$ and $v(x) = \sin x$. [By the Quotient Rule of Differentiation, $\frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right] = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}$].

Step 2: Determining Individual Derivatives

• Let $u = e^x$. The derivative of the exponential function with respect to $x$ is: $\frac{du}{dx} = e^x$. [Based on the derivative rule for $e^x$].
• Let $v = \sin x$. The derivative of the sine function with respect to $x$ is: $\frac{dv}{dx} = \cos x$. [Based on trigonometric differentiation identities].

Step 3: Applying the Quotient Rule

Substituting the variables into the Quotient Rule formula:

$\frac{d}{dx} \left( \frac{e^x}{\sin x} \right) = \frac{(\sin x) \cdot \frac{d}{dx}(e^x) - (e^x) \cdot \frac{d}{dx}(\sin x)}{(\sin x)^2}$

Step 4: Algebraic Simplification

Substitute the derivatives found in Step 2 into the equation:

$\frac{d}{dx} \left( \frac{e^x}{\sin x} \right) = \frac{(\sin x)(e^x) - (e^x)(\cos x)}{\sin^2 x}$

Factor out the common term $e^x$ from the numerator:

$\frac{d}{dx} \left( \frac{e^x}{\sin x} \right) = \frac{e^x (\sin x - \cos x)}{\sin^2 x}$

Step 5: Final Expression Formulation

The expression can be left in the consolidated form or expanded as follows:

$\frac{e^x \sin x}{\sin^2 x} - \frac{e^x \cos x}{\sin^2 x} = \frac{e^x}{\sin x} - \frac{e^x \cos x}{\sin^2 x}$

Using trigonometric identities ($\csc x = \frac{1}{\sin x}$ and $\cot x = \frac{\cos x}{\sin x}$):

$\frac{d}{dx} \left( \frac{e^x}{\sin x} \right) = e^x \csc x - e^x \csc x \cot x = e^x \csc x (1 - \cot x)$

Final Solution: $\frac{e^x(\sin x - \cos x)}{\sin^2 x}$ or $e^x \csc x (1 - \cot x)$

Step 1: Establishing the Chain Rule

To differentiate the composite function $f(x) = \sin(\tan^{-1}(e^{-x}))$, we apply the Chain Rule of differentiation. The Chain Rule states that for a composite function $y = f(g(h(x)))$, the derivative is given by:

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

[Per the Chain Rule of Differentiation]

Step 2: Decomposition of Functions

Let $y = \sin(u)$, where $u = \tan^{-1}(v)$, and $v = e^{-x}$. We differentiate each layer sequentially:

• $\frac{dy}{du} = \frac{d}{du}(\sin u) = \cos u$
• $\frac{du}{dv} = \frac{d}{dv}(\tan^{-1} v) = \frac{1}{1 + v^2}$ [Standard derivative of inverse trigonometric functions]
• $\frac{dv}{dx} = \frac{d}{dx}(e^{-x}) = -e^{-x}$ [Chain Rule applied to exponential functions]

Step 3: Execution of the Differentiation

Substituting the variables back into the derivative expression:

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

Simplifying the denominator: $(e^{-x})^2 = e^{-2x}$.

$\frac{dy}{dx} = \frac{-e^{-x} \cos(\tan^{-1}(e^{-x}))}{1 + e^{-2x}}$

Step 4: Geometric Representation of the Argument

To express $\cos(\tan^{-1}(e^{-x}))$ in algebraic form, let $\theta = \tan^{-1}(e^{-x})$, which implies $\tan \theta = e^{-x} = \frac{e^{-x}}{1}$.

Based on the triangle, the hypotenuse is $\sqrt{1^2 + (e^{-x})^2} = \sqrt{1 + e^{-2x}}$. Thus, $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{1 + e^{-2x}}}$.

Step 5: Final Simplification

Substitute $\cos(\tan^{-1}(e^{-x})) = \frac{1}{\sqrt{1 + e^{-2x}}}$ into the expression from Step 3:

$\frac{dy}{dx} = \frac{-e^{-x}}{1 + e^{-2x}} \cdot \frac{1}{\sqrt{1 + e^{-2x}}}$

$\frac{dy}{dx} = \frac{-e^{-x}}{(1 + e^{-2x})^{3/2}}$

Final Solution: $\frac{dy}{dx} = \frac{-e^{-x}}{(1 + e^{-2x})^{3/2}}$

Step 1: Formal Representation of the Function

Let the given function be represented by $y(x)$:

$y = e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5}$

Step 2: Applying the Linearity Property of Differentiation

To find the derivative $\frac{dy}{dx}$, we apply the Sum Rule of differentiation, which states that the derivative of a sum of functions is the sum of their individual derivatives [Per the Linearity of the Differentiation Operator]:

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

Step 3: Application of the Chain Rule

For composite functions of the form $e^{f(x)}$, we apply the Chain Rule [By the Chain Rule: $\frac{d}{dx}(e^{u}) = e^u \cdot \frac{du}{dx}$]:

• $\frac{d}{dx}(e^x) = e^x \cdot \frac{d}{dx}(x) = e^x \cdot 1 = e^x$
• $\frac{d}{dx}(e^{x^2}) = e^{x^2} \cdot \frac{d}{dx}(x^2) = e^{x^2} \cdot (2x) = 2x e^{x^2}$
• $\frac{d}{dx}(e^{x^3}) = e^{x^3} \cdot \frac{d}{dx}(x^3) = e^{x^3} \cdot (3x^2) = 3x^2 e^{x^3}$
• $\frac{d}{dx}(e^{x^4}) = e^{x^4} \cdot \frac{d}{dx}(x^4) = e^{x^4} \cdot (4x^3) = 4x^3 e^{x^4}$
• $\frac{d}{dx}(e^{x^5}) = e^{x^5} \cdot \frac{d}{dx}(x^5) = e^{x^5} \cdot (5x^4) = 5x^4 e^{x^5}$

Step 4: Summation of Individual Derivatives

Substituting the results from Step 3 back into the summation expression:

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

Visual Representation of the Power Gradient

Final Solution: $\frac{dy}{dx} = e^x + 2x e^{x^2} + 3x^2 e^{x^3} + 4x^3 e^{x^4} + 5x^4 e^{x^5}$

Step 1: Identifying the Function and Rule

Let the given function be denoted as $y$:

$y = \log(\cos e^x)$

To differentiate this function with respect to $x$, we must apply the Chain Rule of differentiation, which 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]

Step 2: Decomposition of Composite Layers

We decompose the function into three distinct nested functions:

• Outer function: $f(u) = \log(u)$, where $u = \cos(e^x)$
• Middle function: $g(v) = \cos(v)$, where $v = e^x$
• Inner function: $h(x) = e^x$

Step 3: Sequential Differentiation

Applying the derivative $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$:

1. Derivative of outer function: $\frac{d}{du}(\log u) = \frac{1}{u}$

2. Derivative of middle function: $\frac{d}{dv}(\cos v) = -\sin v$

3. Derivative of inner function: $\frac{d}{dx}(e^x) = e^x$ [Per the standard derivative formula for exponential functions]

Step 4: Algebraic Synthesis

Substituting the components back into the product chain:

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

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

Step 5: Simplification using Trigonometric Identities

We observe the term $\frac{\sin(e^x)}{\cos(e^x)}$, which is defined as $\tan(e^x)$ [Per the quotient identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$]:

$\frac{dy}{dx} = - \left( \frac{\sin(e^x)}{\cos(e^x)} \right) \cdot e^x$

$\frac{dy}{dx} = -e^x \tan(e^x)$

Final Solution: $\frac{dy}{dx} = -e^x \tan(e^x)$

Given Variables & Initial Setup

We are required to find the derivative of the function $f(x)$ with respect to $x$, where:

$f(x) = \frac{\cos x}{\log x}$

Given the condition $x > 0$ and $x \neq 1$ (since $\log 1 = 0$, which would make the denominator undefined), we apply the calculus of quotients.

Step 1: Application of the Quotient Rule

To differentiate a quotient of two functions $u(x)$ and $v(x)$, we employ the Quotient Rule [Calculus: Differentiation of Quotients]:

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

Here, let:

• $u(x) = \cos x \implies u'(x) = -\sin x$ [By trigonometric differentiation rule]
• $v(x) = \log x \implies v'(x) = \frac{1}{x}$ [By logarithmic differentiation rule]

Step 2: Substitution into the Formula

Substituting the identified components into the quotient rule expression:

$\frac{d}{dx} \left( \frac{\cos x}{\log x} \right) = \frac{(\log x) \cdot \frac{d}{dx}(\cos x) - (\cos x) \cdot \frac{d}{dx}(\log x)}{(\log x)^2}$

Step 3: Algebraic Simplification

Executing the derivatives determined in Step 1:

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

To eliminate the complex fraction in the numerator, we multiply the numerator and denominator by $x$:

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

Step 4: Final Formatting

Factoring out the negative sign from the numerator for clarity:

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

Final Solution: $\frac{d}{dx} \left( \frac{\cos x}{\log x} \right) = -\frac{x \sin x \log x + \cos x}{x(\log x)^2}$

Step 1: Establishing the Function

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

$y = \log(\log x)$

The domain is given as $x > 1$. Since $\log x$ must be positive for the outer logarithm to be defined, we require $\log x > 0$, which implies $x > 1$. This satisfies the existence condition of the composite function [Per the definition of logarithmic domain constraints].

Step 2: Applying the Chain Rule

To differentiate a composite function of the form $f(g(x))$, we use the Chain Rule [By the Chain Rule of Differentiation]:

$\frac{dy}{dx} = \frac{d}{dx} [\log(\log x)]$

Let $u = \log x$. Then the expression becomes $y = \log(u)$.

According to the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Step 3: Calculating Individual Derivatives

First, differentiate the outer function with respect to $u$ [Using the standard derivative formula $\frac{d}{dx}(\log x) = \frac{1}{x}$]:

$\frac{dy}{du} = \frac{d}{du}(\log u) = \frac{1}{u}$

Second, differentiate the inner function $u$ with respect to $x$:

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

Step 4: Substitution and Simplification

Substitute the derivatives back into the chain rule product:

$\frac{dy}{dx} = \left( \frac{1}{u} \right) \cdot \left( \frac{1}{x} \right)$

Replace $u$ with its original definition in terms of $x$ ($u = \log x$):

$\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{1}{x}$

$\frac{dy}{dx} = \frac{1}{x \log x}$

Visualization of the Derivative Process

Final Solution: $\frac{dy}{dx} = \frac{1}{x \log x}$

Step 1: Identification of the Function Structure

Given the function $y = e^{x^3}$, we identify this as a composite function of the form $f(g(x))$, where:

• Outer function: $f(u) = e^u$
• Inner function: $u = g(x) = x^3$

[By the Chain Rule of differentiation, $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$]

Step 2: Applying the Chain Rule

To differentiate $y = e^{x^3}$ with respect to $x$, we apply the differentiation operator $\frac{d}{dx}$ to both sides:

$\frac{dy}{dx} = \frac{d}{dx}(e^{x^3})$

First, differentiate the outer exponential function with respect to the exponent $x^3$, which yields the exponential function itself [Since $\frac{d}{du}(e^u) = e^u$]:

$\frac{dy}{dx} = e^{x^3} \cdot \frac{d}{dx}(x^3)$

Step 3: Differentiation of the Inner Function

Now, compute the derivative of the inner function $x^3$ using the Power Rule [$\frac{d}{dx}(x^n) = nx^{n-1}$]:

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

Step 4: Algebraic Consolidation

Substitute the result from Step 3 back into the expression derived in Step 2:

$\frac{dy}{dx} = e^{x^3} \cdot (3x^2)$

Rearranging the terms for standard mathematical notation:

$\frac{dy}{dx} = 3x^2 e^{x^3}$

Final Solution: \frac{dy}{dx} = 3x^2 e^{x^3}

Step 1: Identifying the Functional Structure

The given function is a composite function of the form $y = f(g(x))$, where $f(u) = \cos(u)$ and $u = g(x) = \log x + e^x$. To differentiate this, we must employ the Chain Rule of Differentiation [Per the Chain Rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$].

Step 2: Applying the Chain Rule

Let $y = \cos(\log x + e^x)$. We differentiate with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx} [\cos(\log x + e^x)]$

Applying the derivative of the outer function $\cos(u)$, which is $-\sin(u)$:

$\frac{dy}{dx} = -\sin(\log x + e^x) \cdot \frac{d}{dx} (\log x + e^x)$ [By the Chain Rule]

Step 3: Differentiating the Inner Function

We now differentiate the inner expression $(\log x + e^x)$ term-by-term using the Sum Rule of Differentiation [Per the Sum Rule: $\frac{d}{dx}[u + v] = \frac{du}{dx} + \frac{dv}{dx}$]:

• Derivative of $\log x$ is $\frac{1}{x}$ [By standard derivative formula: $\frac{d}{dx}(\ln x) = \frac{1}{x}$]
• Derivative of $e^x$ is $e^x$ [By standard derivative formula: $\frac{d}{dx}(e^x) = e^x$]

Therefore:

$\frac{d}{dx} (\log x + e^x) = \left( \frac{1}{x} + e^x \right)$

Step 4: Final Substitution and Simplification

Substituting the result from Step 3 into the equation from Step 2:

$\frac{dy}{dx} = -\sin(\log x + e^x) \cdot \left( \frac{1}{x} + e^x \right)$

Rearranging the terms for mathematical standard form:

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

Visual Representation of the Derivative Process

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

Step 1: Formal Definition of the Function

Let the given function be represented as $y$:

$y = e^{\sin^{-1} x}$

[Definition of the function for differentiation]

Step 2: Identifying the Differentiation Rule

To differentiate this function, we employ the Chain Rule of calculus. Since the function is a composite function of the form $f(g(x))$, where $f(u) = e^u$ and $g(x) = \sin^{-1} x$, the derivative is given by:

$\frac{dy}{dx} = \frac{d}{du}(e^u) \cdot \frac{d}{dx}(\sin^{-1} x)$

[Per the Chain Rule for differentiation]

Step 3: Execution of the Chain Rule

Applying the standard derivative formulas for the exponential function and the inverse trigonometric function:

1. The derivative of $e^u$ with respect to $u$ is $e^u$.

2. The derivative of $\sin^{-1} x$ with respect to $x$ is $\frac{1}{\sqrt{1 - x^2}}$.

[Standard derivatives: $\frac{d}{dx}e^x = e^x$ and $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$]

Step 4: Algebraic Synthesis

Substituting the derivatives back into the Chain Rule expression:

$\frac{dy}{dx} = e^{\sin^{-1} x} \cdot \frac{d}{dx}(\sin^{-1} x)$

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

Step 5: Visual Representation of the Differential Geometry

Step 6: Final Algebraic Refinement

Combining the terms into a single rational expression:

$\frac{dy}{dx} = \frac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}}$

[Subject to the domain constraint $|x| < 1$ for the derivative to be defined]

Final Solution: $\frac{dy}{dx} = \frac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}}$

Step 1: Identifying the Function and Applying Chain Rule

Let the given function be defined as:

$y = \sqrt{e^{\sqrt{x}}}$

Rewrite the function using exponential notation to facilitate differentiation:

$y = (e^{\sqrt{x}})^{1/2} = e^{\frac{1}{2}\sqrt{x}}$ [By Law of Exponents: $(a^m)^n = a^{mn}$]

Step 2: Applying the Chain Rule of Differentiation

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

Let $u = \frac{1}{2}\sqrt{x} = \frac{1}{2}x^{1/2}$. Then $y = e^u$.

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

Step 3: Calculating Derivative Components

Part A: Differentiation of the exponential term

$\frac{d}{du}(e^u) = e^u$ [Standard derivative formula for exponential functions].

Substituting $u$ back, we get: $e^{\frac{1}{2}\sqrt{x}}$.

Part B: Differentiation of the exponent term

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

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

$\frac{1}{2} \cdot \left(\frac{1}{2}x^{1/2 - 1}\right) = \frac{1}{4}x^{-1/2} = \frac{1}{4\sqrt{x}}$

Step 4: Combining the Components

Multiplying the results from Part A and Part B:

$\frac{dy}{dx} = e^{\frac{1}{2}\sqrt{x}} \cdot \frac{1}{4\sqrt{x}}$

Substitute $e^{\frac{1}{2}\sqrt{x}}$ back to the original radical form $\sqrt{e^{\sqrt{x}}}$:

$\frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}$

Final Solution: $\frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}$