Generating Secure PDF Link

Calculus S.M Yusuf Chapter 2 Ex 2.2 Solved Notes & Quiz

Please Wait

seconds remaining...

← Return to Notes View

Calculus S.M Yusuf Chapter 2 Ex 2.2 Solved Notes & Quiz

Official Academic Study Notes • Published: October 16, 2026 • Free Printable Resource

What are the Core Formulas and Definitions for this Exercise?

In Chapter 2, Exercise 2.2 of Calculus with Analytic Geometry by S.M. Yusuf, we explore advanced techniques of differentiation. These include implicit differentiation, trigonometric substitutions, logarithmic differentiation, and derivative calculation for parametric functions.

Key Mathematical Definitions:

Implicit Function: An equation of the form $F(x, y) = 0$ where $y$ cannot be explicitly isolated as $y = f(x)$. To find $\frac{dy}{dx}$, we differentiate both sides with respect to $x$ term-by-term applying the chain rule to $y$-terms.
Parametric Equations: When $x$ and $y$ are both expressed in terms of a third variable (parameter) $t$ or $\theta$ as $x = f(t)$ and $y = g(t)$, the derivative is found via the chain rule: $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \quad \text{where } \frac{dx}{dt} \neq 0$$
Logarithmic Differentiation: Applied to variable exponents of the form $y = [f(x)]^{g(x)}$. Taking the natural logarithm $\ln y = g(x) \ln[f(x)]$ simplifies the function before differentiating.

Rule / Operation	Standard Formula	Conditions / Notes
Derivative of $\sin^{-1} u$	$\frac{d}{dx}(\sin^{-1} u) = \frac{1}{\sqrt{1-u^2}} \frac{du}{dx}$	$\|u\| < 1$
Derivative of $\cos^{-1} u$	$\frac{d}{dx}(\cos^{-1} u) = -\frac{1}{\sqrt{1-u^2}} \frac{du}{dx}$	$\|u\| < 1$
Derivative of $\tan^{-1} u$	$\frac{d}{dx}(\tan^{-1} u) = \frac{1}{1+u^2} \frac{du}{dx}$	$u \in \mathbb{R}$
Parametric Chain Rule	$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$	$\frac{dx}{d\theta} \neq 0$
Logarithmic Identity	$\frac{d}{dx} \left[ (f(x))^{g(x)} \right] = (f(x))^{g(x)} \frac{d}{dx} \left[ g(x) \ln f(x) \right]$	$f(x) > 0$

How to Solve All Exercise Questions Step-by-Step?

Question 1 (Part i): Find $\frac{dy}{dx}$ for the implicit relation $x^3 + y^3 = 3axy$.

Solution:
Given equation: $$x^3 + y^3 = 3axy$$ Differentiating both sides with respect to $x$: $$\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = 3a \frac{d}{dx}(xy)$$ Using power rule and product rule: $$3x^2 + 3y^2 \frac{dy}{dx} = 3a \left( x \frac{dy}{dx} + y \cdot 1 \right)$$ Divide the entire equation by 3: $$x^2 + y^2 \frac{dy}{dx} = ax \frac{dy}{dx} + ay$$ Group all terms containing $\frac{dy}{dx}$ on the left-hand side: $$y^2 \frac{dy}{dx} - ax \frac{dy}{dx} = ay - x^2$$ Factoring out $\frac{dy}{dx}$: $$(y^2 - ax) \frac{dy}{dx} = ay - x^2$$ $$\frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax}$$
Answer: $$\frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax}$$

Question 1 (Part ii): Find $\frac{dy}{dx}$ for the implicit relation $x^2 + 2hxy + y^2 = 1$.

Solution:
Given equation: $$x^2 + 2hxy + y^2 = 1$$ Differentiating both sides with respect to $x$: $$\frac{d}{dx}(x^2) + 2h \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(1)$$ $$2x + 2h \left( x \frac{dy}{dx} + y \right) + 2y \frac{dy}{dx} = 0$$ Divide by 2: $$x + hx \frac{dy}{dx} + hy + y \frac{dy}{dx} = 0$$ Rearrange terms to collect $\frac{dy}{dx}$: $$(hx + y) \frac{dy}{dx} = -(x + hy)$$ $$\frac{dy}{dx} = -\frac{x + hy}{hx + y}$$
Answer: $$\frac{dy}{dx} = -\frac{x + hy}{hx + y}$$

Question 2 (Part i): Differentiate $y = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ with respect to $x$.

Solution:
Let $x = \tan \theta$, which implies $\theta = \tan^{-1} x$. Substitute $x = \tan \theta$ into $y$: $$y = \tan^{-1}\left(\frac{2\tan \theta}{1 - \tan^2 \theta}\right)$$ Using the double-angle identity for tangent, $\tan(2\theta) = \frac{2\tan \theta}{1 - \tan^2 \theta}$: $$y = \tan^{-1}(\tan 2\theta) = 2\theta$$ Substitute back $\theta = \tan^{-1} x$: $$y = 2 \tan^{-1} x$$ Now, differentiate with respect to $x$: $$\frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1} x) = 2 \cdot \frac{1}{1 + x^2}$$
Answer: $$\frac{dy}{dx} = \frac{2}{1 + x^2}$$

Question 2 (Part ii): Differentiate $y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ with respect to $x$.

Solution:
Let $x = \tan \theta$, so $\theta = \tan^{-1} x$. Substitute $x = \tan \theta$: $$y = \sin^{-1}\left(\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}\right)$$ Recall the cosine double-angle identity: $\cos(2\theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$. $$y = \sin^{-1}(\cos 2\theta)$$ Using the co-function identity $\cos(2\theta) = \sin\left(\frac{\pi}{2} - 2\theta\right)$: $$y = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - 2\theta\right)\right) = \frac{\pi}{2} - 2\theta$$ Substitute $\theta = \tan^{-1} x$: $$y = \frac{\pi}{2} - 2 \tan^{-1} x$$ Differentiating with respect to $x$: $$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) - 2 \frac{d}{dx}(\tan^{-1} x) = 0 - 2 \cdot \frac{1}{1 + x^2} = -\frac{2}{1 + x^2}$$
Answer: $$\frac{dy}{dx} = -\frac{2}{1 + x^2}$$

Question 3 (Part i): Find $\frac{dy}{dx}$ if $y = x^x$.

Solution:
Given function has variable base and variable exponent: $$y = x^x$$ Take natural logarithm on both sides: $$\ln y = \ln(x^x) = x \ln x$$ Differentiating implicitly with respect to $x$: $$\frac{d}{dx}(\ln y) = \frac{d}{dx}(x \ln x)$$ $$\frac{1}{y} \frac{dy}{dx} = x \cdot \frac{1}{x} + \ln x \cdot 1$$ $$\frac{1}{y} \frac{dy}{dx} = 1 + \ln x$$ Multiply by $y$: $$\frac{dy}{dx} = y (1 + \ln x)$$ Substitute back $y = x^x$: $$\frac{dy}{dx} = x^x (1 + \ln x)$$
Answer: $$\frac{dy}{dx} = x^x (1 + \ln x)$$

Question 3 (Part ii): Find $\frac{dy}{dx}$ if $y = (\sin x)^{\cos x}$.

Solution:
Take natural logarithm on both sides: $$\ln y = \ln\left((\sin x)^{\cos x}\right) = \cos x \ln(\sin x)$$ Differentiate both sides with respect to $x$: $$\frac{1}{y} \frac{dy}{dx} = \cos x \cdot \frac{d}{dx}(\ln(\sin x)) + \ln(\sin x) \cdot \frac{d}{dx}(\cos x)$$ $$\frac{1}{y} \frac{dy}{dx} = \cos x \cdot \frac{1}{\sin x} \cdot \cos x + \ln(\sin x) \cdot (-\sin x)$$ $$\frac{1}{y} \frac{dy}{dx} = \frac{\cos^2 x}{\sin x} - \sin x \ln(\sin x) = \cot x \cos x - \sin x \ln(\sin x)$$ Multiply by $y$: $$\frac{dy}{dx} = (\sin x)^{\cos x} \left[ \cot x \cos x - \sin x \ln(\sin x) \right]$$
Answer: $$\frac{dy}{dx} = (\sin x)^{\cos x} \left[ \cot x \cos x - \sin x \ln(\sin x) \right]$$

Question 4 (Part i): Find $\frac{dy}{dx}$ for parametric equations $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$.

Solution:
First, differentiate $x$ with respect to $\theta$: $$\frac{dx}{d\theta} = a \cdot 3 \cos^2 \theta \cdot \frac{d}{d\theta}(\cos \theta) = -3a \cos^2 \theta \sin \theta$$ Next, differentiate $y$ with respect to $\theta$: $$\frac{dy}{d\theta} = a \cdot 3 \sin^2 \theta \cdot \frac{d}{d\theta}(\sin \theta) = 3a \sin^2 \theta \cos \theta$$ Apply the parametric chain rule: $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta}$$ Simplify by canceling $3a \sin \theta \cos \theta$: $$\frac{dy}{dx} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta$$
Answer: $$\frac{dy}{dx} = -\tan \theta$$

Question 4 (Part ii): Find $\frac{dy}{dx}$ for the cycloid equations $x = a(\theta - \sin \theta)$ and $y = a(1 - \cos \theta)$.

Solution:
Differentiate $x$ with respect to $\theta$: $$\frac{dx}{d\theta} = a (1 - \cos \theta)$$ Differentiate $y$ with respect to $\theta$: $$\frac{dy}{d\theta} = a (0 - (-\sin \theta)) = a \sin \theta$$ Apply parametric chain rule: $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \sin \theta}{a (1 - \cos \theta)} = \frac{\sin \theta}{1 - \cos \theta}$$ Using half-angle trigonometric formulas $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$ and $1 - \cos \theta = 2 \sin^2(\theta/2)$: $$\frac{dy}{dx} = \frac{2 \sin(\theta/2) \cos(\theta/2)}{2 \sin^2(\theta/2)} = \frac{\cos(\theta/2)}{\sin(\theta/2)} = \cot\left(\frac{\theta}{2}\right)$$
Answer: $$\frac{dy}{dx} = \cot\left(\frac{\theta}{2}\right)$$

Question 5: If $y = \sqrt{x + \sqrt{x + \sqrt{x + \dots \infty}}}$, prove that $(2y - 1)\frac{dy}{dx} = 1$.

Solution:
Given infinite series: $$y = \sqrt{x + \sqrt{x + \sqrt{x + \dots \infty}}}$$ Since the nested radical extends infinitely, the expression inside the outer radical is equal to $y$ itself: $$y = \sqrt{x + y}$$ Squaring both sides: $$y^2 = x + y$$ Differentiating implicitly with respect to $x$: $$\frac{d}{dx}(y^2) = \frac{d}{dx}(x) + \frac{d}{dx}(y)$$ $$2y \frac{dy}{dx} = 1 + \frac{dy}{dx}$$ Subtract $\frac{dy}{dx}$ from both sides: $$2y \frac{dy}{dx} - \frac{dy}{dx} = 1$$ Factor out $\frac{dy}{dx}$: $$(2y - 1)\frac{dy}{dx} = 1$$
Answer: $$(2y - 1)\frac{dy}{dx} = 1 \quad \text{(Proved)}$$

Question 6: Differentiate $u = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to $v = \tan^{-1} x$.

Solution:
Let $x = \tan \theta \implies \theta = \tan^{-1} x = v$. First, simplify $u$: $$u = \tan^{-1}\left(\frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta}\right)$$ Since $\sqrt{1+\tan^2 \theta} = \sec \theta$: