How do you find the derivative of #tan(arcsin(x))#?

1 Answer
George C.
Aug 28, 2016

#d/(dx) tan(arcsin(x))= 1/(1-x^2)^(3/2)#

Explanation:

Let #t = arcsin(x)#

Then:

#x = sin(t)#

So:

#tan(arcsin(x)) = tan(t) = sin(t)/cos(t) = x/sqrt(1-x^2)#

So:

#d/(dx) tan(arcsin(x))#

#= d/(dx) (x (1-x^2)^(-1/2))#

#= (1-x^2)^(-1/2) + x*(-1/2)(1-x^2)^(-3/2)*(-2x)#

#= (1-x^2)(1-x^2)^(-3/2) +x^2(1-x^2)^(-3/2)#

#= 1/(1-x^2)^(3/2)#

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