Step 1: Formal Definition of Continuity

A function $f(x)$ is continuous at a point $x = c$ if and only if the following three conditions are satisfied [Per the Definition of Continuity]:

• $f(c)$ is defined.
• $\lim_{x \to c} f(x)$ exists, i.e., $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$.
• $\lim_{x \to c} f(x) = f(c)$.

Step 2: Identifying the Domain

The given function is $f(x) = \frac{1}{x - 5}$. The function is a rational function defined for all $x \in \mathbb{R}$ except where the denominator equals zero [By the definition of the domain of rational functions]. Setting the denominator to zero: $x - 5 = 0 \implies x = 5$. Thus, the domain of $f$ is $D_f = \mathbb{R} \setminus \{5\}$.

Step 3: Analytical Verification of Continuity

Let $c$ be any arbitrary real number in the domain such that $c \neq 5$. We must evaluate $\lim_{x \to c} f(x)$:

$\lim_{x \to c} f(x) = \lim_{x \to c} \left( \frac{1}{x - 5} \right)$

Using the Quotient Law for limits, since $\lim_{x \to c} (x - 5) = c - 5 \neq 0$:

$\lim_{x \to c} f(x) = \frac{\lim_{x \to c} (1)}{\lim_{x \to c} (x - 5)} = \frac{1}{c - 5}$

Since $f(c) = \frac{1}{c - 5}$, it is evident that $\lim_{x \to c} f(x) = f(c)$. Therefore, $f(x)$ is continuous for all $c \in \mathbb{R} \setminus \{5\}$.

Step 4: Visual Representation

Step 5: Conclusion on Discontinuity

At $x = 5$, the function is undefined as division by zero is prohibited in the set of real numbers [By the Field Axioms of $\mathbb{R}$]. Because the function is not defined at $x=5$, it fails the first condition of continuity. Consequently, the function has a vertical asymptote at $x = 5$.

Final Solution: The function $f(x) = \frac{1}{x-5}$ is continuous for all $x \in \mathbb{R} \setminus \{5\}$ and is discontinuous at $x = 5$.

Step 1: Differentiation of $x$ with respect to $t$

Given $x = \frac{\sin^3 t}{\sqrt{\cos 2t}}$. We apply the Quotient Rule, $\frac{d}{dt}\left[\frac{u}{v}\right] = \frac{v \cdot u' - u \cdot v'}{v^2}$ [Per the Quotient Rule of Calculus].

Let $u = \sin^3 t \implies \frac{du}{dt} = 3\sin^2 t \cos t$ [By Chain Rule].

Let $v = (\cos 2t)^{1/2} \implies \frac{dv}{dt} = \frac{1}{2}(\cos 2t)^{-1/2} \cdot (-\sin 2t) \cdot 2 = \frac{-\sin 2t}{\sqrt{\cos 2t}}$.

$\frac{dx}{dt} = \frac{\sqrt{\cos 2t} \cdot (3\sin^2 t \cos t) - \sin^3 t \cdot \left( \frac{-\sin 2t}{\sqrt{\cos 2t}} \right)}{\cos 2t}$

$\frac{dx}{dt} = \frac{3\sin^2 t \cos t \cos 2t + \sin^3 t \sin 2t}{(\cos 2t)^{3/2}}$

Using $\sin 2t = 2 \sin t \cos t$:

$\frac{dx}{dt} = \frac{\sin^2 t [3 \cos t \cos 2t + 2 \sin^2 t \cos t]}{(\cos 2t)^{3/2}} = \frac{\sin^2 t \cos t [3 \cos 2t + 2 \sin^2 t]}{(\cos 2t)^{3/2}}$

Step 2: Differentiation of $y$ with respect to $t$

Given $y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$. Applying Quotient Rule:

Let $u = \cos^3 t \implies \frac{du}{dt} = 3\cos^2 t (-\sin t) = -3\cos^2 t \sin t$.

$\frac{dy}{dt} = \frac{\sqrt{\cos 2t} \cdot (-3\cos^2 t \sin t) - \cos^3 t \cdot \left( \frac{-\sin 2t}{\sqrt{\cos 2t}} \right)}{\cos 2t}$

$\frac{dy}{dt} = \frac{-3\cos^2 t \sin t \cos 2t + \cos^3 t \sin 2t}{(\cos 2t)^{3/2}}$

Using $\sin 2t = 2 \sin t \cos t$:

$\frac{dy}{dt} = \frac{-\cos^2 t \sin t [3 \cos 2t - 2 \cos^2 t]}{(\cos 2t)^{3/2}}$

Step 3: Evaluating the Ratio $\frac{dy}{dx}$

Using the Parametric Differentiation Theorem: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ [Per the Chain Rule for Parametric Equations].

$\frac{dy}{dx} = \frac{- \cos^2 t \sin t [3 \cos 2t - 2 \cos^2 t]}{\sin^2 t \cos t [3 \cos 2t + 2 \sin^2 t]}$

Cancel common terms $\sin t$ and $\cos t$:

$\frac{dy}{dx} = -\frac{\cos t [3(2\cos^2 t - 1) - 2\cos^2 t]}{\sin t [3(1 - 2\sin^2 t) + 2\sin^2 t]}$

$\frac{dy}{dx} = -\frac{\cos t [6\cos^2 t - 3 - 2\cos^2 t]}{\sin t [3 - 6\sin^2 t + 2\sin^2 t]} = -\frac{\cos t [4\cos^2 t - 3]}{\sin t [3 - 4\sin^2 t]}$

Step 4: Trigonometric Simplification

Recall the Triple Angle Identities: $\cos 3t = 4\cos^3 t - 3\cos t$ and $\sin 3t = 3\sin t - 4\sin^3 t$.

$\frac{dy}{dx} = -\frac{\cos t (4\cos^2 t - 3)}{\sin t (3 - 4\sin^2 t)} = -\frac{4\cos^3 t - 3\cos t}{3\sin t - 4\sin^3 t}$

$\frac{dy}{dx} = -\frac{\cos 3t}{\sin 3t} = -\cot 3t$

Final Solution: \frac{dy}{dx} = -\cot 3t

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: 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: Analyzing the Parametric Equations

Given the parametric equations for $x$ and $y$ as functions of parameter $t$:

$x = \sqrt{a^{\sin^{-1} t}} = (a^{\sin^{-1} t})^{1/2} = a^{\frac{1}{2}\sin^{-1} t}$

$y = \sqrt{a^{\cos^{-1} t}} = (a^{\cos^{-1} t})^{1/2} = a^{\frac{1}{2}\cos^{-1} t}$

Step 2: Logarithmic Differentiation

To differentiate exponential functions of the form $a^{f(t)}$, we apply the logarithmic derivative property: $\frac{d}{dt}(a^u) = a^u \ln(a) \cdot \frac{du}{dt}$ [Per the Chain Rule and Derivative of Exponential Functions].

Differentiating $x$ with respect to $t$:

$\frac{dx}{dt} = a^{\frac{1}{2}\sin^{-1} t} \cdot \ln(a) \cdot \frac{d}{dt}\left(\frac{1}{2}\sin^{-1} t\right)$

$\frac{dx}{dt} = a^{\frac{1}{2}\sin^{-1} t} \cdot \ln(a) \cdot \frac{1}{2\sqrt{1-t^2}}$

Differentiating $y$ with respect to $t$:

$\frac{dy}{dt} = a^{\frac{1}{2}\cos^{-1} t} \cdot \ln(a) \cdot \frac{d}{dt}\left(\frac{1}{2}\cos^{-1} t\right)$

$\frac{dy}{dt} = a^{\frac{1}{2}\cos^{-1} t} \cdot \ln(a) \cdot \left(-\frac{1}{2\sqrt{1-t^2}}\right)$

Step 3: Applying the Chain Rule for Parametric Differentiation

By the Chain Rule, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ [Per the theorem on parametric differentiation].

$\frac{dy}{dx} = \frac{a^{\frac{1}{2}\cos^{-1} t} \cdot \ln(a) \cdot \left(-\frac{1}{2\sqrt{1-t^2}}\right)}{a^{\frac{1}{2}\sin^{-1} t} \cdot \ln(a) \cdot \left(\frac{1}{2\sqrt{1-t^2}}\right)}$

Step 4: Algebraic Simplification

Cancel the common terms $\ln(a)$, $\frac{1}{2}$, and $\frac{1}{\sqrt{1-t^2}}$:

$\frac{dy}{dx} = -\frac{a^{\frac{1}{2}\cos^{-1} t}}{a^{\frac{1}{2}\sin^{-1} t}}$

Substitute back the original expressions for $x$ and $y$:

Since $y = a^{\frac{1}{2}\cos^{-1} t}$ and $x = a^{\frac{1}{2}\sin^{-1} t}$:

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

Visualization of the Relationship

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

Step 1: Theoretical Definition of Continuity

A function $f(x)$ is continuous at a point $c$ in its domain if and only if:

1. $f(c)$ is defined.
2. $\lim_{x \to c} f(x)$ exists.
3. $\lim_{x \to c} f(x) = f(c)$.

A function is continuous on its entire domain if it is continuous at every point $c$ in its domain [Definition of Continuity on an Interval].

Step 2: Analysis of the Cosine Function: $f(x) = \cos(x)$

The domain of $f(x) = \cos(x)$ is $\mathbb{R}$. For any arbitrary point $c \in \mathbb{R}$:

Consider the limit: $\lim_{x \to c} \cos(x)$. Let $x = c + h$, where $h \to 0$.

$\lim_{h \to 0} \cos(c + h) = \lim_{h \to 0} [\cos(c)\cos(h) - \sin(c)\sin(h)]$ [Per the Trigonometric Identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$].

Since $\lim_{h \to 0} \cos(h) = 1$ and $\lim_{h \to 0} \sin(h) = 0$, we have:

$\cos(c)(1) - \sin(c)(0) = \cos(c)$.

Since $\lim_{x \to c} f(x) = f(c)$, the cosine function is continuous everywhere on $\mathbb{R}$.

Step 3: Analysis of Cosecant, Secant, and Cotangent

These functions are defined as reciprocals of sine and cosine functions. Specifically:

• $\csc(x) = \frac{1}{\sin(x)}$, defined for $\sin(x) \neq 0 \implies x \neq n\pi, n \in \mathbb{Z}$.
• $\sec(x) = \frac{1}{\cos(x)}$, defined for $\cos(x) \neq 0 \implies x \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z}$.
• $\cot(x) = \frac{\cos(x)}{\sin(x)}$, defined for $\sin(x) \neq 0 \implies x \neq n\pi, n \in \mathbb{Z}$.

[Theorem: If $g(x)$ and $h(x)$ are continuous functions, then the quotient $f(x) = \frac{g(x)}{h(x)}$ is continuous at all points where $h(x) \neq 0$.]

Step 4: Summary of Discontinuity Points

Function	Domain	Points of Discontinuity
$\cos(x)$	$\mathbb{R}$	None
$\csc(x)$	$\mathbb{R} - \{n\pi : n \in \mathbb{Z}\}$	$\{n\pi : n \in \mathbb{Z}\}$
$\sec(x)$	$\mathbb{R} - \{(2n+1)\frac{\pi}{2} : n \in \mathbb{Z}\}$	$\{(2n+1)\frac{\pi}{2} : n \in \mathbb{Z}\}$
$\cot(x)$	$\mathbb{R} - \{n\pi : n \in \mathbb{Z}\}$	$\{n\pi : n \in \mathbb{Z}\}$

Step 5: Visual Representation (Unit Circle-based Sine/Cosine continuity)

Final Solution: The cosine function is continuous for all $x \in \mathbb{R}$. The cosecant and cotangent functions are continuous for all $x \in \mathbb{R} \setminus \{n\pi, n \in \mathbb{Z}\}$. The secant function is continuous for all $x \in \mathbb{R} \setminus \{(2n+1)\frac{\pi}{2}, n \in \mathbb{Z}\}$.

Given Variables & Initial Setup

We are tasked with examining the continuity of the function $f(x)$ defined as:

$f(x) = \frac{x^2 - 25}{x+5}, \quad x \neq -5$

To analyze the continuity at any point $c$ in the domain of the function, we must evaluate whether $\lim_{x \to c} f(x) = f(c)$ [Per the formal definition of continuity].

Step 1: Algebraic Simplification

The function is a rational function. We observe the numerator $x^2 - 25$, which is a difference of squares: $a^2 - b^2 = (a-b)(a+b)$.

$f(x) = \frac{(x-5)(x+5)}{x+5}$

Since the domain constraint is explicitly $x \neq -5$, the factor $(x+5)$ can be cancelled from the numerator and the denominator without loss of generality [By the property of limits in rational functions].

$f(x) = x - 5, \quad \forall x \in \mathbb{R} \setminus \{-5\}$

Step 2: Analysis of Continuity

A function is continuous if it is continuous at every point in its domain. Let $c$ be any real number such that $c \neq -5$.

• Calculate $f(c)$: $f(c) = c - 5$.
• Calculate the limit: $\lim_{x \to c} f(x) = \lim_{x \to c} (x - 5) = c - 5$.

Since $\lim_{x \to c} f(x) = f(c)$ for all $c \in (-\infty, -5) \cup (-5, \infty)$, the function $f(x)$ is continuous at all points in its domain.

Visual Representation

Conclusion on Domain Constraint

At $x = -5$, the function $f(x)$ is undefined because the denominator becomes zero. Thus, the function is not continuous at $x = -5$ because it fails the existence condition for continuity [Per the definition of continuity at a point]. However, within the specified domain $D = \{x \in \mathbb{R} : x \neq -5\}$, the function behaves as a linear polynomial, which is continuous everywhere.

Final Solution: The function $f(x) = \frac{x^2 - 25}{x+5}$ is continuous for all $x \in \mathbb{R} \setminus \{-5\}$. It is discontinuous at $x = -5$ as the point is excluded from the domain.

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: Differentiation of $x$ with respect to $t$

Given the parametric equation for $x$: $x = a\left(\cos t + \log \tan\left(\frac{t}{2}\right)\right)$

To find $\frac{dx}{dt}$, apply the linearity of the derivative [Per the Sum Rule and Chain Rule]:

$\frac{dx}{dt} = a \left[ \frac{d}{dt}(\cos t) + \frac{d}{dt}\left(\log \tan\frac{t}{2}\right) \right]$

Calculating the individual components:

• $\frac{d}{dt}(\cos t) = -\sin t$
• $\frac{d}{dt}(\log \tan\frac{t}{2}) = \frac{1}{\tan(t/2)} \cdot \sec^2\left(\frac{t}{2}\right) \cdot \frac{1}{2}$ [By the Chain Rule]

Simplifying the second component using trigonometric identities:

$\frac{1}{\tan(t/2)} \cdot \sec^2\left(\frac{t}{2}\right) \cdot \frac{1}{2} = \frac{\cos(t/2)}{\sin(t/2)} \cdot \frac{1}{\cos^2(t/2)} \cdot \frac{1}{2} = \frac{1}{2 \sin(t/2) \cos(t/2)}$

Using the double angle identity $\sin(2\theta) = 2\sin\theta\cos\theta$, we get:

$\frac{1}{\sin t} = \csc t$

Therefore:

$\frac{dx}{dt} = a(-\sin t + \csc t) = a\left(\frac{1 - \sin^2 t}{\sin t}\right) = a\frac{\cos^2 t}{\sin t}$

Step 2: Differentiation of $y$ with respect to $t$

Given the parametric equation for $y$: $y = a \sin t$

Differentiating with respect to $t$:

$\frac{dy}{dt} = a \frac{d}{dt}(\sin t) = a \cos t$ [By the derivative of trigonometric functions]

Step 3: Applying the Parametric Differentiation Rule

To find $\frac{dy}{dx}$, we use the formula [Per the Parametric Chain Rule]:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

Substituting the derived expressions:

$\frac{dy}{dx} = \frac{a \cos t}{a \frac{\cos^2 t}{\sin t}}$

Step 4: Algebraic Simplification

Cancel common factors $a$ and $\cos t$ (assuming $\cos t \neq 0$):

$\frac{dy}{dx} = \frac{\cos t}{\frac{\cos^2 t}{\sin t}} = \cos t \cdot \frac{\sin t}{\cos^2 t} = \frac{\sin t}{\cos t}$

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

$\frac{dy}{dx} = \tan t$

Final Solution: $\frac{dy}{dx} = \tan t$

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: Definition of the Absolute Value Function

The function is defined as $f(x) = |x - 5|$. By the definition of the modulus function, this can be expressed as a piecewise function:

$f(x) = \begin{cases} -(x - 5) & \text{if } x < 5 \\ x - 5 & \text{if } x \ge 5 \end{cases}$

[Per the algebraic definition of absolute value $|u| = u$ if $u \ge 0$ and $-u$ if $u < 0$].

Step 2: Analysis of Continuity at an Arbitrary Point $c$

To examine the continuity of $f(x)$ for all $x \in \mathbb{R}$, we consider three distinct cases for an arbitrary real number $c$:

Case 1: $c < 5$

In the interval $(-\infty, 5)$, the function is defined by $f(x) = -x + 5$. This is a polynomial function of degree 1 [Per the theorem that every polynomial function is continuous for all real numbers].

$\lim_{x \to c} f(x) = \lim_{x \to c} (-x + 5) = -c + 5 = f(c)$. Thus, $f(x)$ is continuous for all $x < 5$.

Case 2: $c > 5$

In the interval $(5, \infty)$, the function is defined by $f(x) = x - 5$. This is also a polynomial function.

$\lim_{x \to c} f(x) = \lim_{x \to c} (x - 5) = c - 5 = f(c)$. Thus, $f(x)$ is continuous for all $x > 5$.

Case 3: $c = 5$

We must evaluate the left-hand limit (LHL), right-hand limit (RHL), and the value of the function at $x = 5$:

• Value of function: $f(5) = |5 - 5| = 0$.
• Left-hand limit: $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} -(x - 5) = -(5 - 5) = 0$.
• Right-hand limit: $\lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (x - 5) = (5 - 5) = 0$.

[Since $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5)$, the function is continuous at $x = 5$].

Step 3: Visual Representation

Analytical Summary

Point of Analysis	LHL	RHL	$f(c)$	Status
$c < 5$	$-c+5$	$-c+5$	$-c+5$	Continuous
$c = 5$	$0$	$0$	$0$	Continuous
$c > 5$	$c-5$	$c-5$	$c-5$	Continuous

Final Solution: The function $f(x) = |x - 5|$ is continuous for all real numbers $x \in \mathbb{R}$.

Given Variables & Initial Setup

The function $f(x)$ is defined piecewise as:

Domain Interval	Function Definition
$x \le 5$	$f(x) = kx + 1$
$x > 5$	$f(x) = 3x - 5$

For a function $f(x)$ to be continuous at a point $x = c$, the following condition must hold [Per the definition of continuity at a point]:

$$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$$

Step 1: Calculating the Left-Hand Limit (LHL) at $x = 5$

The LHL is evaluated using the definition of the function for $x \le 5$:

$$\text{LHL} = \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (kx + 1)$$

By substituting $x = 5$ into the expression [Direct Substitution Property]:

$$\text{LHL} = k(5) + 1 = 5k + 1$$

Step 2: Calculating the Right-Hand Limit (RHL) at $x = 5$

The RHL is evaluated using the definition of the function for $x > 5$:

$$\text{RHL} = \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (3x - 5)$$

By substituting $x = 5$ into the expression:

$$\text{RHL} = 3(5) - 5 = 15 - 5 = 10$$

Step 3: Calculating the Function Value at $x = 5$

The value of the function at the point of continuity is given by the definition where $x = 5$:

$$f(5) = k(5) + 1 = 5k + 1$$

Step 4: Equating Limits to Determine $k$

For $f(x)$ to be continuous at $x = 5$, the LHL must equal the RHL [Theorem: Continuity at a Point]:

$$5k + 1 = 10$$

Subtract $1$ from both sides:

$$5k = 9$$

Divide by $5$:

$$k = \frac{9}{5}$$

Visual Representation of Continuity at $x=5$

Final Solution: The value of $k$ that ensures the function $f$ is continuous at $x = 5$ is $k = \frac{9}{5}$ (or $1.8$).

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: Simplifying the Inverse Trigonometric Function

Given the equation: $y = \sec^{-1}\left(\frac{1}{2x^2-1}\right)$.

Utilize the trigonometric identity for the reciprocal of the secant function: $\sec^{-1}\left(\frac{1}{u}\right) = \cos^{-1}(u)$ [Per the property of inverse trigonometric functions, where $|u| \geq 1$].

Applying this to the expression:

$y = \cos^{-1}(2x^2 - 1)$

Step 2: Trigonometric Substitution

Given the domain constraint $0 < x < \frac{1}{\sqrt{2}}$, we choose a substitution that simplifies the expression $2x^2 - 1$.

Let $x = \cos \theta$. Then $\theta = \cos^{-1}x$.

Substituting into the expression:

$y = \cos^{-1}(2\cos^2\theta - 1)$

Applying the double-angle formula for cosine: $\cos(2\theta) = 2\cos^2\theta - 1$ [By trigonometric identity]:

$y = \cos^{-1}(\cos 2\theta)$

Step 3: Evaluating the Range and Simplifying

Since $0 < x < \frac{1}{\sqrt{2}}$, it follows that:

$0 < \cos \theta < \frac{1}{\sqrt{2}}$

This implies $\frac{\pi}{4} < \theta < \frac{\pi}{2}$.

Multiplying by 2: $\frac{\pi}{2} < 2\theta < \pi$.

Since $2\theta \in (\frac{\pi}{2}, \pi)$, the expression $\cos^{-1}(\cos 2\theta)$ simplifies to $2\theta$ [By the property $\cos^{-1}(\cos \alpha) = \alpha$ for $\alpha \in [0, \pi]$].

Substituting $\theta = \cos^{-1}x$ back into the equation:

$y = 2\cos^{-1}x$

Step 4: Differentiation

Now, differentiate with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx} [2\cos^{-1}x]$

Applying the constant multiple rule and the derivative formula $\frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}$:

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

Step 5: Final Result

The derivative is computed as follows:

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

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

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

Given Variables & Initial Setup

We are provided with the parametric equations for $x$ and $y$ as functions of the parameter $\theta$:

• $x = a(\cos \theta + \theta \sin \theta)$
• $y = a(\sin \theta - \theta \cos \theta)$

To find $\frac{dy}{dx}$, we utilize the chain rule for parametric differentiation: $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ [Per the Rule of Parametric Differentiation].

Step 1: Differentiating $x$ with respect to $\theta$

Given $x = a(\cos \theta + \theta \sin \theta)$, we differentiate with respect to $\theta$ using the Sum Rule and Product Rule:

$\frac{dx}{d\theta} = \frac{d}{d\theta} [a(\cos \theta + \theta \sin \theta)]$

$\frac{dx}{d\theta} = a \left[ \frac{d}{d\theta}(\cos \theta) + \frac{d}{d\theta}(\theta \sin \theta) \right]$

Applying the product rule $\frac{d}{d\theta}(uv) = u\frac{dv}{d\theta} + v\frac{du}{d\theta}$ to the term $\theta \sin \theta$:

$\frac{dx}{d\theta} = a [-\sin \theta + (\theta \cos \theta + \sin \theta \cdot 1)]$

$\frac{dx}{d\theta} = a [-\sin \theta + \theta \cos \theta + \sin \theta]$

$\frac{dx}{d\theta} = a \theta \cos \theta$

Step 2: Differentiating $y$ with respect to $\theta$

Given $y = a(\sin \theta - \theta \cos \theta)$, we differentiate with respect to $\theta$:

$\frac{dy}{d\theta} = \frac{d}{d\theta} [a(\sin \theta - \theta \cos \theta)]$

$\frac{dy}{d\theta} = a \left[ \frac{d}{d\theta}(\sin \theta) - \frac{d}{d\theta}(\theta \cos \theta) \right]$

Applying the product rule to the term $\theta \cos \theta$:

$\frac{dy}{d\theta} = a [\cos \theta - (\theta (-\sin \theta) + \cos \theta \cdot 1)]$

$\frac{dy}{d\theta} = a [\cos \theta - (-\theta \sin \theta + \cos \theta)]$

$\frac{dy}{d\theta} = a [\cos \theta + \theta \sin \theta - \cos \theta]$

$\frac{dy}{d\theta} = a \theta \sin \theta$

Step 3: Calculating $\frac{dy}{dx}$

Substitute the results from Step 1 and Step 2 into the parametric derivative formula:

$\frac{dy}{dx} = \frac{a \theta \sin \theta}{a \theta \cos \theta}$

By canceling the common terms $a$ and $\theta$ (where $\theta \neq 0$):

$\frac{dy}{dx} = \frac{\sin \theta}{\cos \theta}$

$\frac{dy}{dx} = \tan \theta$

Final Solution: $\frac{dy}{dx} = \tan \theta$

Given Equation & Differentiation Strategy

The given equation is:

$ax + by^2 = \cos y$

To find $\frac{dy}{dx}$, we must perform implicit differentiation with respect to $x$ across the entire equation [Per the Chain Rule of Calculus].

Step 1: Differentiating Both Sides with Respect to $x$

Applying the operator $\frac{d}{dx}$ to both sides of the equation:

$\frac{d}{dx}(ax + by^2) = \frac{d}{dx}(\cos y)$

Step 2: Applying Sum and Constant Multiple Rules

Distributing the derivative across the terms and factoring out constants:

$a \cdot \frac{d}{dx}(x) + b \cdot \frac{d}{dx}(y^2) = \frac{d}{dx}(\cos y)$

[By the Linearity Property of Differentiation]

Step 3: Executing the Chain Rule

Differentiate each term individually, keeping in mind that $y$ is a function of $x$ ($y=f(x)$):

• $\frac{d}{dx}(ax) = a(1) = a$
• $b \cdot \frac{d}{dx}(y^2) = b \cdot (2y \cdot \frac{dy}{dx}) = 2by \frac{dy}{dx}$ [By Power Rule and Chain Rule]
• $\frac{d}{dx}(\cos y) = -\sin y \cdot \frac{dy}{dx}$ [By Derivative of Trigonometric Functions and Chain Rule]

Substituting these back into the equation:

$a + 2by \frac{dy}{dx} = -\sin y \frac{dy}{dx}$

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

Group all terms containing $\frac{dy}{dx}$ on one side of the equation:

$a = -\sin y \frac{dy}{dx} - 2by \frac{dy}{dx}$

$a = -\frac{dy}{dx} (\sin y + 2by)$

Step 5: Solving for $\frac{dy}{dx}$

Divide both sides by the coefficient of $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{a}{ -(\sin y + 2by)}$

$\frac{dy}{dx} = -\frac{a}{2by + \sin y}$

Geometric Representation of Implicit Differentiation

Final Solution: $\frac{dy}{dx} = -\frac{a}{2by + \sin y}$

Step 1: Analyzing the Trigonometric Substitution

Given the expression $y = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$, we recognize that the interior fraction resembles the double-angle identity for sine: $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$.

To simplify the derivative, we perform the substitution $x = \tan\theta$, which implies $\theta = \tan^{-1}(x)$.

[Justification: Per trigonometric identity $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$].

Step 2: Transforming the Function

Substituting $x = \tan\theta$ into the equation:

$y = \sin^{-1} \left( \frac{2\tan\theta}{1+\tan^2\theta} \right)$
$y = \sin^{-1} (\sin 2\theta)$
$y = 2\theta$

[Justification: Since $\sin^{-1}(\sin \alpha) = \alpha$ for $\alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}]$].

Step 3: Re-substituting $x$ and Differentiating

Replacing $\theta$ with $\tan^{-1}(x)$:

$y = 2\tan^{-1}(x)$

Now, we differentiate both sides with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx} [2\tan^{-1}(x)]$
$\frac{dy}{dx} = 2 \cdot \frac{d}{dx} [\tan^{-1}(x)]$

[Justification: By the Constant Multiple Rule of differentiation].

Step 4: Applying Standard Derivative Formula

Using the standard derivative formula for inverse trigonometric functions:

$\frac{d}{dx} (\tan^{-1} x) = \frac{1}{1+x^2}$
$\frac{dy}{dx} = 2 \cdot \left( \frac{1}{1+x^2} \right)$

[Justification: Differentiation of inverse tangent function].

Step 5: Visual Representation of the Differentiation Geometry

Derivative Step	Result
Substitution	$x = \tan\theta$
Simplified Function	$y = 2\tan^{-1}(x)$
Resulting Derivative	$\frac{2}{1+x^2}$

Final Solution: \frac{dy}{dx} = \frac{2}{1+x^2}

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: Identifying the Parametric System

We are given two equations where $x$ and $y$ are functions of a common parameter $t$:

• $x = 2at^2$
• $y = at^4$

[Per the Chain Rule for parametric differentiation, if $x = f(t)$ and $y = g(t)$, then $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, provided $\frac{dx}{dt} \neq 0$.]

Step 2: Differentiating $x$ with respect to $t$

Applying the Power Rule of differentiation ($\frac{d}{dt}[t^n] = nt^{n-1}$):
$\frac{dx}{dt} = \frac{d}{dt}(2at^2)$
$\frac{dx}{dt} = 2a \cdot \frac{d}{dt}(t^2)$
$\frac{dx}{dt} = 2a \cdot (2t)$
$\frac{dx}{dt} = 4at$

Step 3: Differentiating $y$ with respect to $t$

Applying the Power Rule of differentiation:
$\frac{dy}{dt} = \frac{d}{dt}(at^4)$
$\frac{dy}{dt} = a \cdot \frac{d}{dt}(t^4)$
$\frac{dy}{dt} = a \cdot (4t^3)$
$\frac{dy}{dt} = 4at^3$

Step 4: Calculating $\frac{dy}{dx}$

Using the relation $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:
$\frac{dy}{dx} = \frac{4at^3}{4at}$
Canceling the common terms $4a$ (assuming $a \neq 0$):
$\frac{dy}{dx} = \frac{t^3}{t}$
Applying the Quotient Rule for exponents ($\frac{t^m}{t^n} = t^{m-n}$):
$\frac{dy}{dx} = t^{3-1}$
$\frac{dy}{dx} = t^2$

Final Solution: $\frac{dy}{dx} = t^2$