# Thread: using the chain rule on for trigonometric functions

1. ## using the chain rule on for trigonometric functions

i have little to no idea what's going on here after the first step. Can someone explain? Im not sure what to ask beyond that.

thanks

2. ## Re: using the chain rule on for trigonometric functions

Originally Posted by kingsolomonsgrave

i have little to no idea what's going on here after the first step. Can someone explain? Im not sure what to ask beyond that.

thanks
The first step was to let \displaystyle \displaystyle \begin{align*} u = \csc{x} + \tan^2{2x} \end{align*} so that \displaystyle \displaystyle \begin{align*} f = \sin{u} \end{align*}. Then \displaystyle \displaystyle \begin{align*} \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \end{align*}. Finding \displaystyle \displaystyle \begin{align*} \frac{df}{du} \end{align*} is easy. In this case, \displaystyle \displaystyle \begin{align*} \frac{du}{dx} \end{align*} is not so easy, you have to use the Chain Rule again. See how you go.