I am having some trouble with some questions I am solving. I am aware they are simple; as I used to solve them without proof back in high school. Now since I am studying Mathematics I need to prove their work.

Question is: Use the definition of a derivative to calculate the derivative of $f(x)=x^2\cos x$ at $x=0$.

I know the answer is $f'(x)=2x\cos x-x^2\sin x$. I just can't prove it.

I know I have to use $f'(x)=lim x-> 0$ $(f(x)-f(0))/(x-0)$ since that is the definition. I get $f'(x)=\frac{x^2\cos x-0}x$ by doing that.

How do I go on with this question?

  • $\begingroup$ $f'(x)=(f(x)-f(0))/(x-0)$is wrong $\endgroup$ – user Mar 17 '18 at 11:04

You are on the right track. You just missed the limit part.

When you are calculating the derivative of a function $f(x)$, you don't just calculate $f'(x)|_{x=x_0}=\frac{f(x_0+h)-f(x_0)}{(x_0+h)-(x_0)}$. Rather, you have to compute the following limit:

$$f'(x)|_{x=x_0}=\color{red}{\lim_\limits{h\to 0}}\frac{f(x_0+h)-f(x_0)}{(x_0+h)-(x_0)}$$

In this case, $f(x)=x^2 \cos x$

Hence, the derivative will be $$f'(x)|_{x=0}=\lim_\limits{h\to 0}\frac{f(0+h)-f(0)}{(0+h)-0}$$ $$=\lim_\limits{h\to 0}\frac{h^2 \cos h - 0}{h}$$ $$=\lim_\limits{h\to 0}(h \cos h)$$ $$=\left(\lim_\limits{h\to 0}h\right) \cdot \left(\lim_\limits{h\to 0}\cos h \right)$$ $$=0 \times 1 = 0$$

This is totally as per the definition of a derivative.

Hope this helps you.


this is $$\frac{f(x)}{x}=x\cos(x)$$ and note that $$|x\cos(x)|\le |x|$$ and this tends to Zero if $x$ tends to zero


Note that at the point $x=0$

$$f'(0)=\lim_{x\to 0}\frac{x^2\cos x-0}{x-0}$$

and more in general at the point $x=x_0$

$$f'(x_0)=\lim_{x\to x_0}\frac{x^2\cos x-x_0^2\cos x_0}{x-x_0}$$

  • $\begingroup$ I already said that in my post. I can't get beyond that. I know the answer is $f'(x)=2x\cos x-x^2\sin x$ for this question. I just cant derive that to the answer. $\endgroup$ – Arszilla Mar 17 '18 at 11:00
  • $\begingroup$ @Arszilla if you are interested to the derivative at a specific point you need to calculate the imit at that point, if you want derive the general expression for every x it is a bit different. $\endgroup$ – user Mar 17 '18 at 11:02
  • $\begingroup$ @Arszilla Note also that in the OP you have not mentioned the limit, and the derivative is defined by a limit. $\endgroup$ – user Mar 17 '18 at 11:04
  • $\begingroup$ I mean the question has product rule and the definition according to my instructor is $f'(x)=(f(x)-f(a))/(x-a)$ for finding derivatives at a certain point. He even solved $f(x)=x^2$ at $x=1$ with this method. $\endgroup$ – Arszilla Mar 17 '18 at 11:06
  • $\begingroup$ @Arszilla please check this definition there is something wrong with that what we can say is as an approximation for $x\approx a$ $$f'(a)\approx(f(x)-f(a))/(x-a)$$ but the definition of derivative need limits, that pretty sure! $\endgroup$ – user Mar 17 '18 at 11:08


We know that $$\frac {d}{dx} ( h(x).\cdot g(x))=h'(x) g(x)+h(x)g'(x)$$ Hence with $h(x)=x^2$ and $g(x)=\cos x$

The $$f'(x)=2x\cos x-x^2\sin x$$

Because $$h'(x)=2x$$ and $$g'(x)=-\sin x$$

Now hence $f'(0)=0$

  • $\begingroup$ Is that the definition? I am not sure if I should use the product rule or the normal way aka $f'(x)=lim x-> 0$ $(f(x)-f(0))/(x-0)$ $\endgroup$ – Arszilla Mar 17 '18 at 11:12
  • $\begingroup$ @Arszilla No it's not the definition. I just used some properties of derivatives. Sorry I didn't knew you need to find the derivative using the major definition $\endgroup$ – Rohan Shinde Mar 17 '18 at 11:14

If I interpret your question correctly, you want to prove that
for two functions $f$ and $g$, $(f\times g)' = [f\times(g')+g\times(f')].$
The following proof is taken verbatim from Calculus, 2nd Ed., vol 1, 1966, by Tom Apostol.

$$f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}.$$ Therefore, $$[f(x)\times g(x)]' = \lim_{h\rightarrow 0} \frac{f(x+h)\times g(x+h) -f(x)\times g(x)}{h}$$

$$ =\;\;\lim_{h\rightarrow 0}\left\{ g(x)\times \frac{f(x+h) - f(x)}{h} + f(x+h)\times\frac{g(x+h) - g(x)}{h}\right\}.$$

Since $f'$ is presumed to exist at $x, f$ is continuous at $x.$Therefore, as $h\rightarrow 0, f(x+h)\rightarrow f(x).\;$ Therefore, the above limit may be re-expressed as

$$ =\;\;\lim_{h\rightarrow 0}\left\{ g(x)\times \frac{f(x+h) - f(x)}{h} + f(x)\times\frac{g(x+h) - g(x)}{h}\right\}$$ $$ = g(x)\times f'(x) + f(x)\times g'(x).$$


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