# If f(x) =cos3x  and g(x) = (2x-1)^2 , what is f'(g(x)) ?

Mar 21, 2016

$\frac{\mathrm{df}}{\mathrm{dg}} = \frac{- 3 \sin 3 x}{4 \left(2 x - 1\right)}$

#### Explanation:

As per chain formula $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dx}}$. hence

$\frac{\mathrm{df}}{\mathrm{dg}} = \frac{\frac{\mathrm{df}}{\mathrm{dx}}}{\frac{\mathrm{dg}}{\mathrm{dx}}}$

As $f \left(x\right) = \sin 3 x$ and $g \left(x\right) = {\left(2 x - 1\right)}^{2}$

$\frac{\mathrm{df}}{\mathrm{dx}} = - \sin 3 x \times 3$ and $\frac{\mathrm{dg}}{\mathrm{dx}} = 2 \left(2 x - 1\right) \times 2$

Hence $\frac{\mathrm{df}}{\mathrm{dg}} = \frac{- 3 \sin 3 x}{4 \left(2 x - 1\right)}$