Question

differentiation of sinx^cos3x

Answer 1

we know a formula :
d/dx(u^v) = u^v × d/dx (v.logu)
in this case we have u=sinx and v=cos3x.
now ,
d/dx(sinx^cos3x)=sinx^cosx ×[d/dx{cos3x .log(sinx)}] =sin^cos3x.[ cos3x.d/dx(logsinx)+log sinx.d/dx(cos3x)]
=sinx^cos3x.[cos3xcosx/sinx+3logsinx.cosx]
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Answer 2

We know that

d/dx(u^v) = u^v × d/dx (v.logu)

We have u=  sinx and v = cos3x.

Now,

d/dx(sinx^cos3x) = sinx^cosx ×[d/dx{cos3x .log(sinx)}] = sin^cos3x.[ cos3x.d/dx(logsinx) + log sinx.d/dx(cos3x)] = sinx^cos3x.[cos3xcosx/sinx+3logsinx.cosx]