How do you derive #y = (sinx)/[1 - sin^2(x))# using the quotient rule?

1 Answer
Gió
Nov 30, 2015

I found: #y'=(cos(x)[1+sin^2(x)])/([1-sin^2(x)]^2)=#

Explanation:

Given a function as #f(x)=(h(x))/g(x)# the Quotient Rule tells us that:

#y'=f'(x)=(h'(x)g(x)-h(x)g'(x))/([g(x)]^2)#

In your case you get:

#y'=f´(x)=(cos(x)(1-sin^2(x))-sin(x)(-2sin(x)cos(x)))/([1-sin^2(x)]^2)=#
#=(cos(x)-cos(x)sin^2(x)+2sin^2(x)cos(x))/([1-sin^2(x)]^2)=#
#=(cos(x)+cos(x)sin^2(x))/([1-sin^2(x)]^2)=#
#=(cos(x)[1+sin^2(x)])/([1-sin^2(x)]^2)=#

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