Question

lf $f\left(x\right)=log\left(\frac{1+x}{1-x}\right)$ and $g\left(x\right)=\frac{3x+x^3}{1+3x^2}$ then $\frac{d}{dx}\left(f\right(g\left(x\right)\left)\right)$ equals

• A. $-f^{{}^{\prime }}\left(x\right)$
• B. $-f\left(x\right)$
• C. $3\left(f\right(x){)}^2$
• D. $3f^{{}^{\prime }}\left(x\right)$

Answer 1

Answer: (D)

$3f^{{}^{\prime }}\left(x\right)$

Lets put the expression on g(x) in place of x in f(x) we can get f(g(x)).
So $f\left(g\right(x\left)\right)=log[1+3x+3x^2+x^3/1+3x^2-3x-x^3]$
$f\left(g\right(x\left)\right)=log\left[\right(1+x{)}^3/(1-x{)}^3]=3f\left(x\right)$
$f‘\left(g\right(x\left)\right)=3f‘\left(x\right)$
Option D