# How do you differentiate f(x)=sin(1/(3x-1)) using the chain rule?

May 14, 2016

$\frac{- 3 \cos \left(\frac{1}{3 x - 1}\right)}{3 x - 1} ^ 2$

#### Explanation:

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

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

$f \left(g \left(x\right)\right) = \sin \left(\frac{1}{3 x - 1}\right) \Rightarrow f ' \left(g \left(x\right)\right) = \cos \left(\frac{1}{3 x - 1}\right)$

and$g \left(x\right) = \frac{1}{3 x - 1} = {\left(3 x - 1\right)}^{-} 1$

$\Rightarrow g ' \left(x\right) = - 1 {\left(3 x - 1\right)}^{-} 2 .3 = - 3 {\left(3 x - 1\right)}^{-} 2 = \frac{- 3}{3 x - 1} ^ 2$
$\text{-----------------------------------------------------}$
Substitute these values into$\textcolor{red}{\text{ (A)}}$

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