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
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".