Here's the question you clicked on:
Idealist
Find the second derivative of f(x)=sinxcosx.
Well, what's the first derivative?
use the product rule...
The first derivative is (cosx)^2-(sinx)^2. But how do I take the derivative of that?
well, remember that cos(a+b)=cos a cos b-sin a sin b...
in other words, use the double angle formula for to reduce the following expression you have obtained. And then use the chain rule.
\[f'(x)=cosx(\cos(x))+(-\sin(x)\sin(x))\] 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
bring the power down, keep the inside, differentiate inside
Since no one seems to get it USE COS^2 X-SIN^2 X=COS(2x) TO SIMPLIFY EXPRESSION
simplified expression is dy/dx=cos(2x) then take the dervative of that
@Idealist are you followign?
Yes. I got it. Thanks for the help, guys.
I like medals. LOL but yeah, nice to see :) let me know your final answer so I can check it