# trig derivative

• Oct 13th 2011, 05:04 PM
delgeezee
trig derivative
what it the derivative of $\displaystyle (cosX)^3$ using rules

Is it as easy as $\displaystyle 3(-sinX)2 =-3sin^2X$ ?
• Oct 13th 2011, 05:09 PM
skeeter
Re: trig derivative
Quote:

Originally Posted by delgeezee
what it the derivative of $\displaystyle (cosX)^3$ using rules

Is it as easy as $\displaystyle 3(-sinX)2 =-3sin^2X$ ?

$\displaystyle \frac{d}{dx} (\cos{x})^3 = 3(\cos{x})^2 \cdot (-\sin{x})$
• Oct 13th 2011, 05:11 PM
pickslides
Re: trig derivative
You need to use the chain rule here

Find $\displaystyle (\cos^3x)'$

Let's make $\displaystyle y=\cos^3x$ and $\displaystyle u=\cos x\implies y=u^3$

So $\displaystyle \frac{dy}{du} = 3u^2$ and $\displaystyle \frac{du}{dx} = -\sin x$

By the chain rule $\displaystyle \frac{dy}{dx}= \frac{dy}{du}\times \frac{du}{dx}$

Can you finish it?
• Oct 13th 2011, 05:13 PM
esander4
Re: trig derivative
Chain rule. 3(cosx)^2(-sinx)
• Oct 13th 2011, 05:22 PM
delgeezee
Re: trig derivative
Thanks, I understand now.