Answer to Question #167829 in Calculus for Vishal

Question

Find f'(x) of f(x)= sin²x.sinx².sin²x²

Answer 1

If f₁(x) = sin²x.sinx² and f₂(x) = sin²x² then

f(x) = f₁(x)*f₂(x)

f'(x) = f₁'(x)*f₂(x) + f₁(x)*f₂'(x)= (sin²x.sinx²)’.sin²x²+ sin²x.sinx².(sin²x²)’

f₁'(x) = (sin²x.sinx²)’ = 2sinx cosx sinx²+sin²x 2xcosx² = 2sinx (cosx sinx² +x sinx cosx²)    

f₂'(x) = (sin²x²)’ = 2 sinx² cosx² 2x = 4x sinx² cosx²

f'(x) = 2sinx (cosx sinx² +x sinx cosx²) sin²x²+ 4x sinx² cosx² sin²x sinx² =

= sin²x² [2sinx (cosx sinx² +x sinx cosx²)+ 4x cosx² sin²x] =

= 2sinx sin²x² [cosx sinx² +x sinx cosx²+ 2x cosx² sinx] =

= 2sinx sin²x² [cosx sinx² +3x sinx cosx²]

Answer:

f'(x) = 2sinx sin²x² [cosx sinx² +3x sinx cosx²]