# Trying to work ahead: Derivatives of Trig Functions

• Mar 23rd 2008, 02:26 PM
NAPA55
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! (Yes)
• Mar 23rd 2008, 02:37 PM
Jhevon
Quote:

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?

Quote:

$\displaystyle y = tan^2(x^3)$
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 $\displaystyle g(x) = x^3$
• Mar 23rd 2008, 02:37 PM
Moo
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 ;-)
• Mar 23rd 2008, 02:56 PM
NAPA55
Thanks! I'll run it out like you showed me and I'll see if I come up with the same answer. (Clapping)
• Mar 23rd 2008, 03:12 PM
NAPA55
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?
• Mar 23rd 2008, 03:36 PM
Moo
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 :-)
• Mar 23rd 2008, 04:10 PM
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?
• Mar 23rd 2008, 04:29 PM
Jhevon
Quote:

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)
• Mar 23rd 2008, 04:35 PM
NAPA55
I'm an idiot. The x came from the denominator function LOL