How do you find the derivative of #(cos^2(x)sin^2(x))#?

Question

Ezra Adams · Accepted Answer

#1/2sin4x# #"differentiate using the "color(blue)"product/chain rules"# #"given "y=f(x)g(x)" then"# #dy/dx=f(x)g'(x)+g(x)f'(x)larrcolor(blue)"product rule"# #f(x)=cos^2xrArrf'(x)=2cosx(-sinx)# #color(white)(xxxxxxxxxxxxxxx)=-2sinxcosx# #g(x)=sin^2xrArrg'(x)=2sinx(cosx)# #color(white)(xxxxxxxxxxxxxxx)=2sinxcosx# #d/dx(cos^2xsin^2x)# #=cos^2x(2sinxcosx)+sin^2x(-2sinxcosx)# #=(2sinxcosx)(cos^2x-sin^2x)# #=sin2xcos2x=1/2sin4x#

Paisley Johnson · Answer

undefined To find the derivative of $ (\cos^2(x)\sin^2(x)) $, you would use the product rule. Here are the steps:

1. Identify the functions as u and v: 
   $ u = \cos^2(x) $ and $ v = \sin^2(x) $.

2. Calculate the derivatives of u and v:
   $ u' = -2\cos(x)\sin(x) $ and $ v' = 2\sin(x)\cos(x) $.

3. Apply the product rule formula:
   $ (uv)' = u'v + uv' $.

4. Substitute the derivatives and functions:
   $ (\cos^2(x)\sin^2(x))' = (-2\cos(x)\sin(x))(\sin^2(x)) + (\cos^2(x))(2\sin(x)\cos(x)) $.

5. Simplify the expression to get the final result for $ (\cos^2(x)\sin^2(x))' $.