# How do you find dy/dx by implicit differentiation of sinx=x(1+tany)?

Jan 13, 2017

$y ' \left(x\right) = \frac{x \left(x \cos x - \sin x\right)}{{\sin}^{2} x - 2 x \sin x + 2 {x}^{2}}$

#### Explanation:

Differentiate both sides of the equation with respect to $x$:

$\frac{d}{\mathrm{dx}} \left(\sin x\right) = \frac{d}{\mathrm{dx}} \left(x \left(1 + \tan y\right)\right)$

remembering that:

$\frac{d}{\mathrm{dx}} f \left(y \left(x\right)\right) = \frac{\mathrm{df}}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

So:

$\cos x = \left(1 + \tan y\right) + \frac{x y '}{\cos} ^ 2 y$

Solving for $y '$:

$y ' \left(x\right) = \left(\cos x - 1 - \tan y\right) {\cos}^{2} \frac{y}{x}$

From the original equation:

$\sin x = x \left(1 + \tan y\right)$

we have that:

$\tan y = \sin \frac{x}{x} - 1$

and as:

${\cos}^{2} y = \frac{1}{\sec} ^ 2 y = \frac{1}{1 + {\tan}^{2} y}$

${\cos}^{2} y = \frac{1}{1 + {\left(\sin \frac{x}{x} - 1\right)}^{2}} = {x}^{2} / \left({\sin}^{2} x - 2 x \sin x + 2 {x}^{2}\right)$

We can make $y ' \left(x\right)$ explicit in $x$ using these expressions:

$y ' \left(x\right) = \left(\cos x - 1 - \sin \frac{x}{x} + 1\right) {x}^{2} / \left({\sin}^{2} x - 2 x \sin x + 2 {x}^{2}\right) = \frac{x \left(x \cos x - \sin x\right)}{{\sin}^{2} x - 2 x \sin x + 2 {x}^{2}}$