# How do you find the derivative of y=sin^2x/cosx?

Jan 5, 2016

$y ' = \frac{2 {\cos}^{2} x \sin x + {\sin}^{3} x}{\cos} ^ 2 x$

#### Explanation:

Use the quotient rule, which states that for a function $y = \frac{f \left(x\right)}{g \left(x\right)}$,

$y ' = \frac{f ' \left(x\right) g \left(x\right) - g ' \left(x\right) f \left(x\right)}{g \left(x\right)} ^ 2$

We have:

$f \left(x\right) = {\sin}^{2} x$
$g \left(x\right) = \cos x$

To find $f ' \left(x\right)$, use the chain rule: $\frac{d}{\mathrm{dx}} \left({u}^{2}\right) = 2 u \cdot u '$.

$f ' \left(x\right) = 2 \sin x \cos x$
$g ' \left(x\right) = - \sin x$

Plug these into the quotient rule equation.

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

$y ' = \frac{2 {\cos}^{2} x \sin x + {\sin}^{3} x}{\cos} ^ 2 x$

This alternatively can be simplified as

$y ' = {\sin}^{3} \frac{x}{\cos} ^ 2 x + 2 \sin x$

Jan 5, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sin x \left(\frac{1}{{\cos}^{2} x} + 1\right)$

#### Explanation:

Note that ${\sin}^{2} x = 1 - {\cos}^{2} x$

Therefore,

$y = \frac{1 - {\cos}^{2} x}{\cos x}$

$= \frac{1}{\cos} x - \cos x$.

Using the Chain Rule,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\frac{1}{\cos} x\right) - \frac{d}{\mathrm{dx}} \left(\cos x\right)$

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

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