Question

If $y=\left(f\right(f\left(x\right)){)}^3,f(0)=0$ and $f^{\prime }\left(0\right)=2,$ then ${\left(\frac{dy}{dx}\right)}_{x=0}$ is equal to

• A. $0$
• B. $2$
• C. $4$
• D. $16$

Answer 1

The correct option is A $0$

Given, $y=\left(f\right(f\left(x\right)){)}^3$
$\Rightarrow \frac{dy}{dx}=3{\left[f\right(f\left(x\right)\left)\right]}^2\cdot f^{\prime }\left(f\right(x\left)\right)\cdot f^{\prime }\left(x\right)\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=0}=3{\left[f\right(0\left)\right]}^2\cdot f^{\prime }\left(0\right)\cdot f^{\prime }\left(0\right)\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=0}=3\cdot 0\cdot 2\cdot 2=0$