Question

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

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

Answer: (A)

$1$

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

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

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

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

$=x$

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