# How do you differentiate  f(x)=1/sqrt((7-2x^3) using the chain rule.?

##### 1 Answer
Apr 14, 2016

$\frac{3 {x}^{2}}{7 - 2 {x}^{3}} ^ \left(\frac{3}{2}\right)$

#### Explanation:

Using the $\textcolor{b l u e}{\text{ chain rule }}$

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

rewrite $\frac{1}{\sqrt{7 - 2 {x}^{3}}} = {\left(7 - 2 {x}^{3}\right)}^{- \frac{1}{2}}$
$\text{-----------------------------------------}$

here f(g(x)) $= {\left(7 - 2 {x}^{3}\right)}^{- \frac{1}{2}}$

$\Rightarrow f ' \left(g \left(x\right)\right) = - \frac{1}{2} {\left(7 - 2 {x}^{3}\right)}^{- \frac{3}{2}}$

and $g \left(x\right) = 7 - 2 {x}^{3} \Rightarrow g ' \left(x\right) = - 6 {x}^{2}$
$\text{----------------------------------------------------------------}$

$\Rightarrow f ' \left(x\right) = - \frac{1}{2} {\left(7 - 2 {x}^{3}\right)}^{- \frac{3}{2}} \times - 6 {x}^{2}$

$= \frac{3 {x}^{2}}{7 - 2 {x}^{3}} ^ \left(\frac{3}{2}\right)$