Question

What is f'(x) of x+sinx÷xcos?

Answer 1

`f'(x)= 1+ (x-sinxcosx)/(x^2cos^2x)`

Explanation:

Assuming `f(x) =x +sinx/(xcosx)`

`f'(x) = 1 + d/dx(sinx/(xcosx))`

Apply quotient rule.

`f'(x) = 1+ (xcosx*cosx - sinx* d/dx(xcosx))/(xcosx)^2`

Apply product rule.

`= 1+ (xcosx*cosx - sinx* (-xsinx+cosx))/(xcosx)^2`

`= 1+ (xcos^2x + xsin^2x - sinxcosx)/(xcosx)^2`

`= 1+ (x(cos^2x + sin^2x) - sinxcosx)/(xcosx)^2`

`= 1+ (x-sinxcosx)/(x^2cos^2x)`