GO Classes 2025 | Weekly Quiz 21 | Calculus | Question: 18

Answer 1

Detailed Video Solution: https://youtu.be/6aIuj2b2J38 

This is one of my favorite questions, a similar kind of question was asked once in my IISc class by Prof Shevade. Surprisingly, very few people were able to provide an answer to it. (Sachin Mittal)

Simple and elegant Question.

Since $g$ is not differentiable we cannot use the product rule.
$$
f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{h g(h)}{h}=8 .
$$

Detailed Video Solution: https://youtu.be/6aIuj2b2J38 

Comments

@Sachin Mittal 1 sir, unknowingly applying product is getting us to correct answer 

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How applying product rule when $g’(0)$ is not given ?

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$f(x)=x.g(x)$

$f'(x)=x.g'(x)+g(x)$

$f'(0)=0.g'(x)+g(0)$

$f'(x)=8$

I know this is wrong and marked option $D$ without giving a thought about 1st principles

But this is that kind of question where Wrong path leads to the correct destination

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yeah was thinking in same direction  but after seeing g is not differentiable fr chod diya  is que

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Details for above explanation: If we think intuitively, differentiation is nothing but slope. Integration is nothing but area under the region.

Since we cannot directly apply differentiation, find slope.

Slope = change in y / change in x

In our case, take two points that are very close to each other. Thus limit comes into picture.