Question:
If ƒ(1) = 1, ƒ'(1) = 3, then the derivative of $f(f(f(\mathrm{x})))+(f(\mathrm{x}))^{2}$ at $\mathrm{x}=1$ is :
Correct Option: , 2
Solution:
$y=f(f(f(\mathrm{x})))+(f(\mathrm{x}))^{2}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=f^{\prime}(f(f(\mathrm{x}))) f^{\prime}(f(\mathrm{x})) f^{\prime}(\mathrm{x})+2 f(\mathrm{x}) f^{\prime}(\mathrm{x})$
$=f^{\prime}(1) f^{\prime}(1) f^{\prime}(1)+2 f(1) f^{\prime}(1)$
$=3 \times 5 \times 3+2 \times 1 \times 3$
$=27+6$
$=33$
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