Question:
If $f(1)=1, f^{\prime}(1)=3$, then the derivative of $f(f(f(x)))+(f(x))^{2}$ at $x=1$ is :
Correct Option: 1
Solution:
Let $g(x)=f(f(f(x)))+(\mathrm{f}(x))^{2}$
Differentiating both sides w.r.t. $x$, we get
$g^{\prime}(x)=f^{\prime}(f(f(x))) f^{\prime}(f(x)) f^{\prime}(x)+2 f(x) f^{\prime}(x)$
$g^{\prime}(1)=f^{\prime}(f(f(1))) f^{\prime}(f(1)) f^{\prime}(1)+2 f(1) f^{\prime}(1)$
$=f^{\prime}(f(1)) \mathrm{f}^{\prime}(1) f^{\prime}(1)+2 f(1) f^{\prime}(1)$
$=3 \times 3 \times 3+2 \times 1 \times 3=27+6=33$