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

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Answer 1 · 2017-06-14T19:38:20

#dy/dx = cotx - sinx cosx#

#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#