Question

If $y=fofof\left(x\right)$ and $f\left(0\right)=0,f^{\prime }\left(0\right)=2$, then find $y^{\prime }\left(0\right)$

• A. $6$
• B. $7$
• C. $8$
• D. $9$

Answer 1

The correct option is C $8$

$y=f\left(f\right(f\left(x\right)\left)\right)Lett=f\left(x\right)andLetu=f\left(f\right(x\left)\right)=f\left(t\right)\therefore \frac{du}{dx}=\frac{du}{dt}\times \frac{dt}{dx}\left(ByChainRule\right)\therefore y=f\left(u\right)=f\left(f\right(t\left)\right)\therefore \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dt}\times \frac{dt}{dx}\left(ByChainRule\right)\therefore y^{{}^{\prime }}=f^{\prime }\left(u\right)\times f^{\prime }\left(t\right)\times f^{\prime }\left(x\right)Atx=0,f\left(0\right)=0andt=f\left(x\right)=f\left(0\right)=0andu=f\left(t\right)=f\left(0\right)=0\therefore y^{{}^{\prime }}=f^{\prime }\left(u\right)\times f^{\prime }\left(t\right)\times f^{\prime }\left(x\right)\therefore y^{{}^{\prime }}\left(0\right)=f^{\prime }\left(0\right)\times f^{\prime }\left(0\right)\times f^{\prime }\left(0\right)=2\times 2\times 2=8$