Idealist Group Title Find the second derivative of f(x)=sinxcosx. one year ago one year ago

1. inkyvoyd Group Title

Well, what's the first derivative?

2. inkyvoyd Group Title

use the product rule...

3. saifoo.khan Group Title

4. Idealist Group Title

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

5. inkyvoyd Group Title

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

6. hcmathkim Group Title

use the chain rule

7. inkyvoyd Group Title

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

8. Sujay Group Title

$f'(x)=cosx(\cos(x))+(-\sin(x)\sin(x))$ 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

9. hcmathkim Group Title

bring the power down, keep the inside, differentiate inside

10. inkyvoyd Group Title

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

11. inkyvoyd Group Title

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

12. inkyvoyd Group Title

@Idealist are you followign?

13. Idealist Group Title

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

14. inkyvoyd Group Title

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