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Idealist 2 years ago Find the second derivative of f(x)=sinxcosx.

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1. inkyvoyd

Well, what's the first derivative?

2. inkyvoyd

use the product rule...

3. saifoo.khan

4. Idealist

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

5. inkyvoyd

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

6. hcmathkim

use the chain rule

7. inkyvoyd

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

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

bring the power down, keep the inside, differentiate inside

10. inkyvoyd

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

11. inkyvoyd

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

12. inkyvoyd

@Idealist are you followign?

13. Idealist

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

14. inkyvoyd

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

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