# If f(x)= csc7 x  and g(x) = e^(1 +3x ) , how do you differentiate f(g(x))  using the chain rule?

Mar 28, 2017

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = - 21 {e}^{\left(1 + 3 x\right)} \csc \left(7 {e}^{1 + 3 x}\right) \cot \left(7 {e}^{1 + 3 x}\right)$

#### Explanation:

As $f \left(x\right) = \csc 7 x$ and $g \left(x\right) = {e}^{1 + 3 x}$

$f \left(g \left(x\right)\right) = \csc \left(7 {e}^{1 + 3 x}\right)$

The chain in the derivative of $f \left(g \left(x\right)\right)$ is in fact longer.

We have $f \left(x\right) = \csc \left(u \left(x\right)\right)$, $u \left(g \left(x\right)\right) = 7 g \left(x\right)$, $g \left(x\right) = {e}^{v \left(x\right)}$ and $v \left(x\right) = 1 + 3 x$.

Hence $\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = \frac{\mathrm{df}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dv}} \times \frac{\mathrm{dv}}{\mathrm{dx}}$

= $- \csc u \cot u \times 7 \times {e}^{v} \times 3$

= $- 21 {e}^{\left(1 + 3 x\right)} \csc \left(7 {e}^{1 + 3 x}\right) \cot \left(7 {e}^{1 + 3 x}\right)$