What is the derivative of y=e^square root of 1+tan(sinx)?

[susan__t]

Chain rule difficulties, due tomorrow!

Homework Statement

Find the derivative of

y=e^square root of 1+tan(sinx)

Homework Equations

chain rule: F'(x)=f'(g(x)) * g'(x)

The Attempt at a Solution

I thought I had it and then while I was looking at other chain rules and started doubting my actual ability to sort out the chain rule...

y=e^square root of 1+tan(sinx)

y'=e^square root of 1+tan(sinx) *(1+tansinx)'

y'=e^square root of 1+tan(sinx)* (0 +sec²sinx +cosx)

Please help! My assignment is due tomorrow and I know there is something not quite right but I don't know why.

 

[jhicks]

how would you evaluate these operations if you wanted to get a number out of them? Remember to take the chain rule in reverse of this order. For example, \frac{d[(x^{4})^3]}{dx} = 3(x^{4})^{2}*4x^{3}dx

ok LaTeX appears to be acting up. Basically you forgot to take the derivative of the square root *before* finding the derivative of what was inside it.

 

[susan__t]

yikes I should probably work on my recopying skills. thank you very much!