ANSWER 1
Note that,
[tex]f(u)=tan^{-1}(u)[/tex]
is the same as
[tex]f(u)=arctan(u)[/tex]
We apply the product rule.
[tex]f(x)=x^2tan^{-1}(x)[/tex]
So we keep the second function and differentiate the first,plus we keep the first function and differentiate the second.
[tex]f'(x)=(x^2)'tan^{-1}(x)+x^2(tan^{-1}(x))' [/tex]
Recall that,
If
[tex]f(u)=tan^{-1}(u)[/tex]
Then,
[tex]f'(u)=\frac{1}{1+u^2}} \times u'[/tex]
This implies that,
[tex]f'(x)=2xtan^{-1}(x)+\frac{x^2}{x^2+1} [/tex]
ANSWER 2
We apply the product rule and the chain rules of differentiation here.
[tex]f(x)=xsin^{-1}(1-x^2)[/tex]
[tex]f'(x)=x'sin^{-1}(1-x^2)+x(sin^{-1}(1-x^2))' [/tex]
Recall that,
If
[tex]f(u)=sin^{-1}(u)[/tex]
Then,
[tex]f'(u)=\frac{1}{\sqrt{1-u^2}} \times u'[/tex]
This implies that,
[tex]f'(x)=sin^{-1}(1-x^2)+x \times \frac{1}{\sqrt{1-(1-x^2)^2}}\times (-2x) [/tex]
[tex]f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{1-(1-2x^2+x^4)}} [/tex]
[tex]f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{1-1+2x^2-x^4}}[/tex]
[tex]f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{2x^2-x^4}}[/tex]
