Exercise 5.3 Practice

Given: The implicit equation is $4x + 5y = \sin x$.

Step 1: Differentiate w.r.t. x
$\frac{d}{dx}(4x) + \frac{d}{dx}(5y) = \frac{d}{dx}(\sin x)$
$4 + 5\frac{dy}{dx} = \cos x$

Step 2: Isolate $\frac{dy}{dx}$
$5\frac{dy}{dx} = \cos x - 4$

Step 1: Differentiate w.r.t. x
$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin y)$
$2 + 3\frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}$ (Using Chain Rule for $\sin y$)

Step 2: Group $\frac{dy}{dx}$ terms
$2 = \cos y \frac{dy}{dx} - 3\frac{dy}{dx}$
$2 = \frac{dy}{dx}(\cos y - 3)$

Step 1: Differentiate w.r.t. x
$\frac{d}{dx}(ax) + \frac{d}{dx}(by^2) = \frac{d}{dx}(\cos y)$
$a + b(2y)\frac{dy}{dx} = -\sin y \frac{dy}{dx}$

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

Step 1: Differentiate w.r.t. x
Use Product Rule for $xy$:
$(x\frac{dy}{dx} + y \cdot 1) + 2y\frac{dy}{dx} = \sec x \tan x + \frac{dy}{dx}$

Step 2: Group $\frac{dy}{dx}$ terms
$x\frac{dy}{dx} + 2y\frac{dy}{dx} - \frac{dy}{dx} = \sec x \tan x - y$
$\frac{dy}{dx}(x + 2y - 1) = \sec x \tan x - y$

Step 1: Differentiate w.r.t. x
$2x + (x\frac{dy}{dx} + y) + 2y\frac{dy}{dx} = 0$

Step 2: Collect terms
$x\frac{dy}{dx} + 2y\frac{dy}{dx} = -(2x + y)$
$\frac{dy}{dx}(x + 2y) = -(2x + y)$

Step 1: Differentiate term by term
$3x^2 + (x^2\frac{dy}{dx} + y(2x)) + (x(2y)\frac{dy}{dx} + y^2(1)) + 3y^2\frac{dy}{dx} = 0$

Step 2: Arrange terms
$3x^2 + 2xy + y^2 + (x^2 + 2xy + 3y^2)\frac{dy}{dx} = 0$

Step 1: Chain Rule
$\frac{d}{dx}(\sin y)^2 + \frac{d}{dx}(\cos(xy)) = \frac{d}{dx}(\pi)$
$2\sin y \cos y \frac{dy}{dx} - \sin(xy) \frac{d}{dx}(xy) = 0$
$\sin 2y \frac{dy}{dx} - \sin(xy) [x\frac{dy}{dx} + y] = 0$

Step 2: Isolate $\frac{dy}{dx}$
$\frac{dy}{dx}(\sin 2y - x\sin(xy)) = y\sin(xy)$

Step 1: Differentiate
$2\sin x \cos x + 2\cos y (-\sin y)\frac{dy}{dx} = 0$
$\sin 2x - \sin 2y \frac{dy}{dx} = 0$

Step 2: Solve
$\sin 2y \frac{dy}{dx} = \sin 2x$

Substitution: Let $x = \tan\theta \Rightarrow \theta = \tan^{-1}x$.

Simplify:
$y = \cos^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right) = \cos^{-1}(\sin 2\theta)$
$y = \cos^{-1}\left(\cos(\frac{\pi}{2} - 2\theta)\right) = \frac{\pi}{2} - 2\theta$
$y = \frac{\pi}{2} - 2\tan^{-1}x$

Differentiate:
$\frac{dy}{dx} = 0 - 2\left(\frac{1}{1+x^2}\right)$

Substitution: Let $x = \tan\theta \Rightarrow \theta = \tan^{-1}x$.

Simplify:
$y = \tan^{-1}\left(\frac{2\tan\theta}{1-\tan^2\theta}\right)$
$y = \tan^{-1}(\tan 2\theta) = 2\theta = 2\tan^{-1}x$

Differentiate:
$\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2}$

Substitution: Let $x = \tan\theta \Rightarrow \theta = \tan^{-1}x$.

Simplify:
$y = \sin^{-1}(\cos 2\theta) = \sin^{-1}\left(\sin(\frac{\pi}{2} - 2\theta)\right)$
$y = \frac{\pi}{2} - 2\tan^{-1}x$

Differentiate:
$\frac{dy}{dx} = -\frac{2}{1+x^2}$

Substitution: Let $x = \sin\theta \Rightarrow \theta = \sin^{-1}x$.

Simplify:
$y = \sin^{-1}(3\sin\theta - 4\sin^3\theta) = \sin^{-1}(\sin 3\theta)$
$y = 3\theta = 3\sin^{-1}x$

Differentiate:
$\frac{dy}{dx} = 3 \cdot \frac{1}{\sqrt{1-x^2}}$

Substitution: Let $x = \sin\theta$. Then $\sqrt{1-x^2} = \cos\theta$.

Simplify:
$y = \cos^{-1}(2\sin\theta\cos\theta) = \cos^{-1}(\sin 2\theta)$
$y = \cos^{-1}(\cos(\frac{\pi}{2} - 2\theta)) = \frac{\pi}{2} - 2\sin^{-1}x$

Differentiate:
$\frac{dy}{dx} = 0 - 2\left(\frac{1}{\sqrt{1-x^2}}\right)$

Substitution: Let $x = \sin\theta$.

Simplify:
$y = \sec^{-1}\left(\frac{1}{1-2\sin^2\theta}\right) = \sec^{-1}\left(\frac{1}{\cos 2\theta}\right)$
$y = \sec^{-1}(\sec 2\theta) = 2\theta = 2\sin^{-1}x$

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

Substitution: Let $x = \tan\theta$.

Simplify:
$y = \tan^{-1}(\tan 3\theta) = 3\theta = 3\tan^{-1}x$

Differentiate:
$\frac{dy}{dx} = 3 \cdot \frac{1}{1+x^2}$