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:)

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- May 31st 2010, 08:16 PMspoc21Trigonometric 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:)

- May 31st 2010, 08:18 PMnikhil
-2(sinx)tan(cosx)[sec(cosx)]^2

- May 31st 2010, 08:55 PMspoc21
- May 31st 2010, 09:50 PMnikhil
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 - Jun 1st 2010, 12:59 AMHallsofIvy
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".