# How do you find the derivative of (sinx^2)/(tanx)?

Mar 3, 2018

$\frac{2 x \cos \left({x}^{2}\right) \tan x - {\sec}^{2} x \sin \left({x}^{2}\right)}{\tan} ^ 2 x$

#### Explanation:

According to the quotient rule, $\left(\frac{f}{g}\right) ' = \frac{f ' g - f g '}{g} ^ 2$

Here, $f = \sin \left({x}^{2}\right)$ and $g = \tan \left(x\right)$

$\frac{d}{\mathrm{dx}} \left(\sin {x}^{2} / \tan x\right) = \frac{\frac{d}{\mathrm{dx}} \left(\sin \left({x}^{2}\right)\right) \cdot \tan x - \frac{d}{\mathrm{dx}} \left(\tan x\right) \cdot \sin \left({x}^{2}\right)}{\tan x} ^ 2$

We first compute:

$\frac{d}{\mathrm{dx}} \left(\sin \left({x}^{2}\right)\right)$

If we take the chain rule, where $f \left(u\right) = \sin \left({x}^{2}\right)$, we have $f = \sin \left(u\right)$ and $u = {x}^{2}$

So we have:

$\frac{d}{\mathrm{du}} \sin \left(u\right) \cdot \frac{d}{\mathrm{dx}} {x}^{2}$

$\cos u \cdot 2 x$

And as $u = {x}^{2}$, we have:

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

Then ,we compute:

$\frac{d}{\mathrm{dx}} \tan x = {\sec}^{2} \left(x\right)$

So we can input into our original equation:

$\frac{2 x \cos \left({x}^{2}\right) \tan x - {\sec}^{2} x \sin \left({x}^{2}\right)}{\tan} ^ 2 x$

Mar 3, 2018

$2 x \cos {x}^{2} \cot x - \sin {x}^{2} {\csc}^{2} x$

#### Explanation:

Let $f \left(x\right) = \sin {x}^{2}$, and $g \left(x\right) = \tan x$

The formula is

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

The derivative of $\sin x$ is $\cos x$, but here we have $\sin {x}^{2}$ so the derivative will be coscolor(red)(x^2 multiplied by the derivative of color(red)(x^2 which is $2 x$, so $2 x \cos {x}^{2}$

So color(blue)(f^'=2xcosx^2

${g}^{'}$ is the derivative of $\tan x$ which is ${\sec}^{2} x$

So color(green)(g^'=sec^2x

Substituting these values in we get:-

${\left(\frac{f}{g}\right)}^{'} = \frac{\textcolor{b l u e}{2 x \cos {x}^{2}} \tan x - \textcolor{g r e e n}{{\sec}^{2} x} \sin {x}^{2}}{\textcolor{b r o w n}{{\tan}^{2} x}}$

Now we just simplify, separate the denominator:-

$\frac{2 x \cos {x}^{2}}{\tan} x - \frac{\textcolor{g r e e n}{\left(\frac{1}{\cos} ^ 2 x\right)} \sin {x}^{2}}{\textcolor{b r o w n}{{\sin}^{2} \frac{x}{\cos} ^ 2 x}}$

$= 2 x \cos {x}^{2} \cot x - \sin {x}^{2} / \cancel{{\cos}^{2} x} \cdot \frac{\cancel{{\cos}^{2} x}}{\sin} ^ 2 x$

$= 2 x \cos {x}^{2} \cot x - \sin {x}^{2} {\csc}^{2} x$