Question

If $f\left(x\right)=\frac{x}{\sqrt{1-x^2}},g\left(x\right)=\frac{x}{\sqrt{1+x^2}}then\frac{d}{dx}\left(fog\right(x\left)\right)=$

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

The correct option is A $1$

$f\left(x\right)=\frac{x}{\sqrt{1-x^2}}$

$g\left(x\right)=\frac{x}{\sqrt{1+x^2}}$

$fog\left(x\right)=f\left(g\right(x\left)\right)$

$=f\left(\frac{x}{\sqrt{1+x^2}}\right)$

$=\frac{\frac{1}{\sqrt{1+x^2}}}{\sqrt{1-\frac{x^2}{1+x^2}}}$

$=\frac{\frac{x}{\sqrt{1+x^2}}}{\frac{\sqrt{1+x^2-x^2}}{\sqrt{1+x^2}}}$

$=\frac{x}{\sqrt{1+x^2-x^2}}$

$=\frac{x}{\sqrt{1}}$

$fog\left(x\right)=x$

$\frac{d}{dx}\left(fog\right(x\left)\right)=\frac{d}{dx}x$

$=1$.