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

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 $\sqrt{1+\tan^2 \theta} = \sec \theta$: