MHT CET · Maths · Differentiation
If \(\mathrm{f}(1)=1, \mathrm{f}^{\prime}(1)=3\), then the derivative of \(\mathrm{f}(\mathrm{f}(\mathrm{f}(x)))+(\mathrm{f}(x))^2\) at \(x=1\) is
- A 12
- B 19
- C 23
- D 33
Answer & Solution
Correct Answer
(D) 33
Step-by-step Solution
Detailed explanation
Let \(y=\mathrm{f}(\mathrm{f}(\mathrm{f}(x)))+(\mathrm{f}(x))^2\)
\(\therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{f}^{\prime}(\mathrm{f}(\mathrm{f}(x))) \cdot \mathrm{f}^{\prime}(\mathrm{f}(x)) \cdot \mathrm{f}^{\prime}(x)+2 \mathrm{f}(x) \mathrm{f}^{\prime}(x)\)
\(\begin{aligned}\left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=1} & =\mathrm{f}^{\prime}(\mathrm{f}(\mathrm{f}(1))) \cdot \mathrm{f}^{\prime}(\mathrm{f}(1)) \cdot \mathrm{f}^{\prime}(1)+2 \mathrm{f}(1) \mathrm{f}^{\prime}(1) \\ & =3 \cdot 3 \cdot 3+2 \cdot 1 \cdot 3 \\ & =33\end{aligned}\)
\(\therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{f}^{\prime}(\mathrm{f}(\mathrm{f}(x))) \cdot \mathrm{f}^{\prime}(\mathrm{f}(x)) \cdot \mathrm{f}^{\prime}(x)+2 \mathrm{f}(x) \mathrm{f}^{\prime}(x)\)
\(\begin{aligned}\left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=1} & =\mathrm{f}^{\prime}(\mathrm{f}(\mathrm{f}(1))) \cdot \mathrm{f}^{\prime}(\mathrm{f}(1)) \cdot \mathrm{f}^{\prime}(1)+2 \mathrm{f}(1) \mathrm{f}^{\prime}(1) \\ & =3 \cdot 3 \cdot 3+2 \cdot 1 \cdot 3 \\ & =33\end{aligned}\)
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