using the chain rule on for trigonometric functions

**kingsolomonsgrave** · October 21st 2012, 07:03 PM

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

**Prove It** · October 21st 2012, 07:06 PM

Originally Posted by kingsolomonsgrave

Click image for larger version.

Name: Picture 15.png
Views: 6
Size: 15.8 KB
ID: 25329

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

Thread: using the chain rule on for trigonometric functions

LinkBack

Thread Tools

Display

using the chain rule on for trigonometric functions

Re: using the chain rule on for trigonometric functions

Similar Math Help Forum Discussions

Chain rule using trig functions

[SOLVED] Chain rule with functions depending on other functions

the chain rule using trigonometric functions

functions using chain rule

Chain rule w/ trigonometric function help

Search Tags