Question

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

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

Answer: (A)

$1$

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

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

$\therefore \frac{d}{dx}\left(fog\right(x\left)\right)=\frac{d}{dx}\left(x\right)=1$