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

##### 3 Answers
Mar 11, 2018

$f ' \left(x\right) = \left(9 {x}^{2} - 2 x\right) \sec \left(3 {x}^{3} - {x}^{2}\right) \tan \left(3 {x}^{3} - {x}^{2}\right)$

#### Explanation:

$f \left(x\right) = y = \sec \left(3 {x}^{3} - {x}^{2}\right)$

chain rule

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

$u = 3 {x}^{3} - {x}^{2} \implies \frac{\mathrm{du}}{\mathrm{dx}} = 9 {x}^{2} - 2 x$

$y = \sec u \implies \frac{\mathrm{dy}}{\mathrm{du}} = \sec u \tan u$

$f ' \left(x\right) = \sec u \tan u \times \left(9 {x}^{2} - 2 x\right)$

$= \left(9 {x}^{2} - 2 x\right) \sec \left(3 {x}^{3} - {x}^{2}\right) \tan \left(3 {x}^{3} - {x}^{2}\right)$

Mar 11, 2018

$\tan \left(3 {x}^{3} - {x}^{2}\right) . \sec \left(3 {x}^{3} - {x}^{2}\right) . x \left(9 x - 2\right)$

#### Explanation:

$y = f \left(g \left(x\right)\right) = \sec \left(3 {x}^{3} - {x}^{2}\right)$

$\frac{\mathrm{df} \left(g \left(x\right)\right)}{\mathrm{dx}} = \frac{\mathrm{df} \left(g \left(x\right)\right)}{\mathrm{dg} \left(x\right)} . \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}}$

$\frac{d \sec \left(3 {x}^{3} - {x}^{2}\right)}{\mathrm{dx}} = \frac{d \sec \left(3 {x}^{3} - {x}^{2}\right)}{d \left(3 {x}^{3} - {x}^{2}\right)} . \frac{d \left(3 {x}^{3} - {x}^{2}\right)}{\mathrm{dx}}$

Because derivative of $\sec \left(x\right)$ is $\tan x . \sec x$ :

$\frac{d \sec \left(3 {x}^{3} - {x}^{2}\right)}{\mathrm{dx}} = \tan \left(3 {x}^{3} - {x}^{2}\right) . \sec \left(3 {x}^{3} - {x}^{2}\right) . 9 {x}^{2} - 2 x$
$= \tan \left(3 {x}^{3} - {x}^{2}\right) . \sec \left(3 {x}^{3} - {x}^{2}\right) . x \left(9 x - 2\right)$

Mar 11, 2018

$f ' \left(x\right) = \left(9 {x}^{2} - 2 x\right) \sec \left(3 {x}^{3} - {x}^{2}\right) \tan \left(3 {x}^{3} - {x}^{2}\right)$

#### Explanation:

â€¢color(white)(x)d/dx(secx)=secxtanx

$\text{Differentiate using the "color(blue)"chain rule}$

$\text{Given "f(x)=g(h(x))" then}$

$f ' \left(x\right) = g ' \left(h \left(x\right)\right) \times h ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

$f \left(x\right) = \sec \left(3 {x}^{3} - {x}^{2}\right)$

$\Rightarrow f ' \left(x\right) = \sec \left(3 {x}^{3} - {x}^{2}\right) \tan \left(3 {x}^{3} - {x}^{2}\right) \times \frac{d}{\mathrm{dx}} \left(3 {x}^{3} - {x}^{2}\right)$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = \left(9 {x}^{2} - 2 x\right) \sec \left(3 {x}^{3} - {x}^{2}\right) \tan \left(3 {x}^{3} - {x}^{2}\right)$