Question

(15 Points)

Answer 1

ANSWER 1

Note that,

$f(u)=tan^{-1}(u)$

is the same as

$f(u)=arctan(u)$

We apply the product rule.

$f(x)=x^2tan^{-1}(x)$

So we keep the second function and differentiate the first,plus we keep the first function and differentiate the second.

$f'(x)=(x^2)'tan^{-1}(x)+x^2(tan^{-1}(x))'$

Recall that,

If

$f(u)=tan^{-1}(u)$

Then,

$f'(u)=\frac{1}{1+u^2}} \times u'$

This implies that,

$f'(x)=2xtan^{-1}(x)+\frac{x^2}{x^2+1}$

ANSWER 2

We apply the product rule and the chain rules of differentiation here.

$f(x)=xsin^{-1}(1-x^2)$

$f'(x)=x'sin^{-1}(1-x^2)+x(sin^{-1}(1-x^2))'$

Recall that,

If

$f(u)=sin^{-1}(u)$

Then,

$f'(u)=\frac{1}{\sqrt{1-u^2}} \times u'$

This implies that,

$f'(x)=sin^{-1}(1-x^2)+x \times \frac{1}{\sqrt{1-(1-x^2)^2}}\times (-2x)$

$f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{1-(1-2x^2+x^4)}}$

$f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{1-1+2x^2-x^4}}$

$f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{2x^2-x^4}}$