How do you differentiate #sin^2(x)#?

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Answer 1 · 2015-10-15T01:27:59

#dy/dx = sin(2x)#

#y = sin^2(x)#

Let's say that #sin(x) = u#, then we have

#y = u^2#

So, if we derivate by #x# we have

#dy/dx = d/dx(u^2)#

By the chain rule #d/dx = d/(du)*(du)/dx#, or

#dy/dx = d/(du)(u^2)*(du)/dx#

#dy/dx = 2u*(du)/dx#

we know #u = sin(x)# so we have

#dy/dx = 2sin(x)*d/dx(sin(x))#

From there we have

#dy/dx = 2sin(x)cos(x)#

Or, if you want, you can squish that into just one formula using the double angle formula for the sine

#dy/dx = sin(2x)#