Question

Question #15d7c

Answer 1

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.