1. ## Trigonometric differentiation help!

Hi, I'm having a lot of trouble solving the following question related to trig differentiation. I would greatly appreciate any help.

1) Differentiate $\displaystyle tan^2 (cos x)$

Thanks

2. -2(sinx)tan(cosx)[sec(cosx)]^2

3. Originally Posted by nikhil
-2(sinx)tan(cosx)[sec(cosx)]^2
Yes, I'm pretty sure thats the answer..Would you mind sharing a bit of the working?

4. d/dx (tan(cosx))^2
= 2tan(cosx)[d/dx(tan(cosx))]
=2tan(cosx)(sec(cosx))^2(d/dx (cosx))
=2tan(cosx)(sec(cosx))^2(-sinx)
=-2(sinx)tan(cosx)(sec(cosx))^2

5. Nikhil used the chain rule.

Let u= cos(x),[/tex]v= tan(u)[/tex], and $\displaystyle w= v^2$

Then $\displaystyle \frac{dw}{dv}= 2v$, $\displaystyle \frac{dv}{du}= sec^2(u)$, and $\displaystyle \frac{du}{dx}= - sin(x)$

The chain rule says
$\displaystyle \frac{dw}{dx}= \frac{dw}{dv}\frac{dv}{du}\frac{du}{dx}$
$\displaystyle = (2v)(sec^2(u))(-sin(x))= (2tan(u))(sec^2(cos(x))(-sin(x))$$\displaystyle = -2tan(cos(x))sec^2(cos(x))sin(x)$.

Of course, nikhil did it with specifically writing out "u", "v", and "w".