Question #15d7c

1 Answer
Oct 11, 2014

Recall that

#(tan^{-1}x)'=1/{1+x^2}#

Let us find the derivative of

#F(x)=tan^{-1}(x/sqrt{1+x^2})#.

(I hope that this is the function you had in mind.)

By Chain Rule,

#F'(x)=1/{1+(x/sqrt{1+x^2})^2}cdot(x/sqrt{1+x^2})'#

by Quotient Rule,

#=1/{1+x^2/{1+x^2}}cdot{1cdot sqrt{1+x^2}-x cdot 1/2(1+x^2)^{-1/2}(2x)}/{1+x^2}#

by simplifying,

#={sqrt{1+x^2}-x^2/sqrt{1+x^2}}/{1+x^2+x^2}#

by multiplying the numerator and the denominator by #sqrt{1+x^2}#,

#={1+x^2-x^2}/{(1+2x^2)sqrt{1+x^2}}#

#=1/{(1+2x^2)sqrt{1+x^2}}#

I hope that this was helpful.