KCET · Maths · Differentiation
If \(f(\mathrm{l})=1, f^{\prime}(\mathrm{l})=3\), then the derivative of \(f(f(f(x)))+(f(x))^2\) at \(x=1\) is
- A 10
- B 33
- C 35
- D 12
Answer & Solution
Correct Answer
(B) 33
Step-by-step Solution
Detailed explanation
Let \(y=f(f(f(x)))+(f(x))^2\)
Differentiating w.r.t. \(x\), we get
\[
\begin{aligned}
& \frac{d y}{d x}=f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x)+2 f(x) \cdot f^{\prime}(x) \\
& \text { At } x=1 \\
& \frac{d y}{d x}=f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1)+2 f(1) \cdot f^{\prime}(1) \\
& \frac{d y}{d x}=f^{\prime}(f(1)) \cdot f^{\prime}(1) \cdot 3+2 \cdot 1 \cdot 3=f^{\prime}(1) \cdot 3 \cdot 3+6 \\
&=9 f^{\prime}(1)+6=9 \cdot 3+6=27+6=33
\end{aligned}
\]
Differentiating w.r.t. \(x\), we get
\[
\begin{aligned}
& \frac{d y}{d x}=f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x)+2 f(x) \cdot f^{\prime}(x) \\
& \text { At } x=1 \\
& \frac{d y}{d x}=f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1)+2 f(1) \cdot f^{\prime}(1) \\
& \frac{d y}{d x}=f^{\prime}(f(1)) \cdot f^{\prime}(1) \cdot 3+2 \cdot 1 \cdot 3=f^{\prime}(1) \cdot 3 \cdot 3+6 \\
&=9 f^{\prime}(1)+6=9 \cdot 3+6=27+6=33
\end{aligned}
\]
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