Find the second derivative of f(x)=sinxcosx.

Question

inkyvoyd · Answer

Well, what's the first derivative?

inkyvoyd · Answer

use the product rule...

saifoo.khan · Answer

attachment

anonymous · Answer

The first derivative is (cosx)^2-(sinx)^2. But how do I take the derivative of that?

inkyvoyd · Answer

well, remember that cos(a+b)=cos a cos b-sin a sin b...

anonymous · Answer

use the chain rule

inkyvoyd · Answer

in other words, use the double angle formula for to reduce the following expression you have obtained.
And then use the chain rule.

anonymous · Answer

$$f'(x)=cosx(\cos(x))+(-\sin(x)\sin(x))$$  Product Rule and Derivatives of Trigs
Simplify.
$$f'(x)=\cos ^{2}x-\sin ^{2}x$$
$$f''(x)=2\cos(x)(-\sin(x))-2\sin(x)(\cos(x))$$  Derivative of Trig and Chain Rule

anonymous · Answer

bring the power down, keep the inside, differentiate inside

inkyvoyd · Answer

Since no one seems to get it
USE COS^2 X-SIN^2 X=COS(2x) TO SIMPLIFY EXPRESSION

inkyvoyd · Answer

simplified expression is dy/dx=cos(2x)
then take the dervative of that

inkyvoyd · Answer

@Idealist are you followign?

anonymous · Answer

Yes. I got it. Thanks for the help, guys.

inkyvoyd · Answer

I like medals.
LOL
but yeah, nice to see :)
let me know your final answer so I can check it