# How do you find the tangent line to x=cos(y/4) at y=π?

Apr 7, 2018

$- 4 \sqrt{2} = \frac{\mathrm{dy}}{\mathrm{dx}}$

#### Explanation:

We have:

$x = \cos \left(\frac{y}{4}\right)$ Let's let $f \left(x\right) = y$.

$\implies x = \cos \left(f \frac{x}{4}\right)$ Apply derivative on both sides.

$\implies \frac{d}{\mathrm{dx}} \left(x\right) = \frac{d}{\mathrm{dx}} \left(\cos \left(f \frac{x}{4}\right)\right)$

$\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$ where $n$ is a constant.

$\frac{d}{\mathrm{dx}} \left(f \left(g \left(x\right)\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

$\frac{d}{\mathrm{dx}} \left(\cos \left(x\right)\right) = - \sin \left(x\right)$

$\implies 1 \cdot {x}^{1 - 1} = - \sin \left(f \frac{x}{4}\right) \cdot \frac{d}{\mathrm{dx}} \left(f \frac{x}{4}\right)$

$\implies {x}^{0} = - \sin \left(f \frac{x}{4}\right) \cdot \frac{1}{4} \cdot \frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)$

$\implies 4 = - \sin \left(f \frac{x}{4}\right) \cdot f ' \left(x\right)$

$\implies \frac{4}{-} \sin \left(f \frac{x}{4}\right) = f ' \left(x\right)$ Remember that $f \left(x\right) = y$ and $f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$

$\implies \frac{4}{-} \sin \left(\frac{y}{4}\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$

When $y = \pi \ldots$

$\implies \frac{4}{-} \sin \left(\frac{\pi}{4}\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$

$\implies \frac{4}{- \frac{\sqrt{2}}{2}} = \frac{\mathrm{dy}}{\mathrm{dx}}$

$\implies 4 \cdot - \frac{2}{\sqrt{2}} = \frac{\mathrm{dy}}{\mathrm{dx}}$

$\implies - \frac{8}{\sqrt{2}} = \frac{\mathrm{dy}}{\mathrm{dx}}$

$\implies - 4 \sqrt{2} = \frac{\mathrm{dy}}{\mathrm{dx}}$