Limit Continuity And Differentiability Ques 34
- If $f(1)=1, f^{\prime}(1)=3$, then the derivative of $f(f(f(x)))+(f(x))^2$ at $x=1$ is
(2019 Main, 8 April II)
(a) $12$
(b) $9$
(c) $15$
(d) $33$
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Answer:
Correct Answer: 34.(d)
Solution: (d) Let $y=f(f(f(x)))+(f(x))^2$
On differentiating both sides w.r.t. $x$, we get
$ \frac{d y}{d x}=f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x)+2 f(x) f^{\prime}(x) $
[by chain rule]
So, $\left.\frac{d y}{d x}\right|_{\text {at } x=1}=f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1)+2 f(1) f^{\prime}(1)$
$\left.\therefore \quad \frac{d y}{d x}\right|_{x=1}=f^{\prime}(f(1)) \cdot f^{\prime}(1) \cdot(3)+2(1)(3)$
$ \left [\because\quad f(1)=1 \text { and } f^{\prime}(1)=3\right] $
$ =f^{\prime}(1) \cdot(3) \cdot(3)+6 $
$ =(3 \times 9)+6=27+6=33$
BETA
