1. ## Trying to work ahead: Derivatives of Trig Functions

Our teacher asked us to look this over this weekend so we can be ahead for next week.

I understood a majority of the questions, and I attempted these two but my answer was quite off.

Could someone help me through these?

Determine the derivative.

$\displaystyle y = \frac{cos2x}{x}$

$\displaystyle y = tan^2(x^3)$

Thanks!

2. Originally Posted by NAPA55
Our teacher asked us to look this over this weekend so we can be ahead for next week.

I understood a majority of the questions, and I attempted these two but my answer was quite off.

Could someone help me through these?

Determine the derivative.

$\displaystyle y = \frac{cos2x}{x}$
what do you fancy? the chain rule or the product rule?

$\displaystyle y = tan^2(x^3)$
use the chain rule here.

$\displaystyle \frac d{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

here, $\displaystyle f(x) = \tan^2 x$ and $\displaystyle g(x) = x^3$

3. Hello,

For the first one, use the formula :

$\displaystyle [\frac{u(x)}{v(x)}]'=\frac{u'(x)v(x)-u(x)v'(x)}{v^2(x)}$

So you should get $\displaystyle \frac{-2x\sin(2x)-\cos(2x)}{x^2}$

For the second one, use :

$\displaystyle [f(g(x))]'=g'(x)*f'(g(x))$

And for this one, write the calculus, you have several ways to find the solution ;-)

4. Thanks! I'll run it out like you showed me and I'll see if I come up with the same answer.

5. Stuck.

Here's what I did:

$\displaystyle f(x) = cos2x = cos^2 x - sin^2 x$

To find the derivative, I tried to apply the product rule twice:
$\displaystyle f '(x) = [-sinxcosx-sinxcosx] - [2sinxcosx]$
$\displaystyle f '(x) = -4sinxcosx$
$\displaystyle f '(x) = -2(2sinxcosx)$
$\displaystyle f '(x) = -2sin2x$

I don't know how you get $\displaystyle -2xsin2x$. Where did the $\displaystyle x$ come from?

6. Well,

Where did the x above go ?

f(x) is not cos(2x), unless there is a typo o.O

Calculating the derivate of cos(2x) is only a step, it's only u'(x). It may have been more simple if you had applied the chain rule :-)

7. I'm still confused as to where that x came from.

Is the derivative of cos2x different than cos(2x)? Are these functions different?

What's the derivative of cos2x?

8. Originally Posted by NAPA55
I'm still confused as to where that x came from.

Is the derivative of cos2x different than cos(2x)? Are these functions different?

What's the derivative of cos2x?
this you would use the chain rule for as well. by the chain rule, the derivative is -2sin(2x)

9. I'm an idiot. The x came from the denominator function LOL