# How do you differentiate f(x)=sqrttan(2-x^3)  using the chain rule?

##### 2 Answers
Dec 2, 2017

f'(x)=-(3x^2sec^2(2-x^3))/(2sqrt(tan(2-x^3))

#### Explanation:

Let $v = - {x}^{3}$ and $u = \tan \left(2 + v\right)$ and $y = f \left(x\right)$

Then $y = \sqrt{u}$.

So $f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dv}} \cdot \frac{\mathrm{dv}}{\mathrm{dx}}$

$\frac{\mathrm{dv}}{\mathrm{dx}} = - 3 {x}^{2}$

$\frac{\mathrm{du}}{\mathrm{dv}} = {\sec}^{2} \left(2 + v\right) = {\sec}^{2} \left(2 - {x}^{3}\right)$

dy/(du)=1/(2sqrtu)=1/(2sqrt(tan(2-x^3))

$\therefore f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = - 3 {x}^{2} \cdot {\sec}^{2} \left(2 - {x}^{3}\right) \cdot \frac{1}{2 \sqrt{\tan \left(2 - {x}^{3}\right)}}$
=-(3x^2sec^2(2-x^3))/(2sqrt(tan(2-x^3))

Dec 2, 2017

f'(x)=(-3x^2sec^2(2-x^3))/(2sqrt(tan(2-x^3))

#### Explanation:

We will use the chain rule where we take the derivative of the outside and multiply it by the derivative of the inside.

We have f(x)=sqrt(tan(2-x^3) It's better to rewrite it as an exponent ${\left(\tan \left(2 - {x}^{3}\right)\right)}^{\frac{1}{2}}$

So, first lets take the derivative of the $\textcolor{b l u e}{\text{outside}}$ using the power rule:

$\textcolor{b l u e}{\frac{1}{2}} \tan {\left(2 - {x}^{3}\right)}^{\textcolor{b l u e}{\frac{1}{2} - \frac{2}{2}}} \implies \frac{1}{2} \tan {\left(2 - {x}^{3}\right)}^{- \frac{1}{2}}$

This is just one part of the answer.

Now let's take the derivative of the $\textcolor{g r e e n}{\text{inside}}$

$\frac{1}{2} {\left(\textcolor{g r e e n}{\tan} \left(2 - {x}^{3}\right)\right)}^{- \frac{1}{2}}$

$\frac{d}{\mathrm{dx}} \textcolor{g r e e n}{\tan} \left(2 - {x}^{3}\right) \implies {\sec}^{2} \left(2 - {x}^{3}\right) \left(- 3 {x}^{2}\right)$

We ended up using the chain rule on the $\textcolor{g r e e n}{\text{inside}}$ too.

The derivative of the $\tan x$ is ${\sec}^{2} x$ but we also have to apply the chain rule here too because ${\sec}^{2} x$ has a function in the inside too. Take the derivative of the $\textcolor{red}{\text{function}}$ inside ${\sec}^{2} \textcolor{red}{\left(2 - {x}^{3}\right)}$ which is where we got the $- 3 {x}^{2}$ from.

Now we just multiply both the derivatives of the outside and inside.

$\frac{1}{2 \sqrt{\tan \left(2 - {x}^{3}\right)}} \times {\sec}^{2} \left(2 - {x}^{3}\right) \left(- 3 {x}^{2}\right)$

d/dxsqrt(tan(2-x^3) $=$ (-3x^2sec^2(2-x^3))/(2sqrt(tan(2-x^3))