Question

Differentiate y=lnsinx-(1/2)sin^(2)x ?

Answer 1

`dy/dx = cotx - sinx cosx`

Explanation:

`y=lnsinx - 1/2 sin^2 x`

The derivative of the `ln` of a function is given by:

`d/dx[lnf(x)] = (f'(x))/(f(x))`

`d/dx(lnsinx) = 1/sinx xx d/dx(sinx) = cosx/sinx=cotx`

The derivative of a function
to a power is:

`d/dx([f(x)]^n) = n[f(x)]^(n-1) f'(x)`

`d/dx(1/2sin^2x) = 2(1/2sinx d/dx(sinx)) = sinx cosx`

`therefore dy/dx = cotx -sinxcosx`