Finding the derivative of a function at an indicated point 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?
 A: 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
A: 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 $x=x_0$
$$f'(x_0)=\lim_{x\to x_0}\frac{x^2\cos x-x_0^2\cos x_0}{x-x_0}$$
A: Hints
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 $f'(0)=0$
A: 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.
A: 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  
$$ =\;\;\lim_{h\rightarrow 0}\left\{
g(x)\times \frac{f(x+h) - f(x)}{h} +
f(x)\times\frac{g(x+h) - g(x)}{h}\right\}$$
$$ = g(x)\times f'(x) + f(x)\times g'(x).$$
