JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Differentiation

Differentiate the two pieces separately. For $f(f(f(x)))$ the chain rule gives $f^{\prime }(f(f(x)))\cdot f^{\prime }(f(x))\cdot f^{\prime }(x)$, and for $(f(x){)}^2$ it gives $2f(x)f^{\prime }(x)$. Now set $x=1$ and use $f(1)=1$, $f^{\prime }(1)=3$. Because $f(1)=1$, every nested evaluation stays at 1, so the first piece becomes $f^{\prime }(1)\cdot f^{\prime }(1)\cdot f^{\prime }(1)=3\times 3\times 3=27$, and the second piece is $2\cdot f(1)\cdot f^{\prime }(1)=2\times 1\times 3=6$. Adding, the derivative at $x=1$ is $27+6=33$.