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I'm trying to help someone with homework and there are quite challenging exercises. I helped her doing most of them but there are one that I want please someone to check and the other I don't have a clue of how to start.

The instructions in these two is:

Use the definition of the derivative to answer the following questions:

  1. If $f(x)$ is a differentiable function, find the derivative of $g(x)=xf(x)$.

  2. If a function satisfies the equation $f(a+b)=f(a)+f(b)+a^2b+ab^2$ for all $a$, $b$ $\in$ $\mathbb{R}$. Suppose $\lim_{x \to 0} \frac {f(x)}{x} = 1$. Find $f(0)$, $f'(0)$ y $f'(x)$.

I start the first one using the definition of derivative $$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$ but I got stucked.

The function is $g(x)$ so:

$$g'(x)=\lim_{h\to 0} \frac{g(x+h)-g(x)}{h}$$

$$g'(x)=\lim_{h\to 0} \frac{(x+h)f(x+h)-xf(x)}{h}$$

$$g'(x)=\lim_{h\to 0} \frac{xf(x+h)+hf(x+h)-xf(x)}{h}$$

$$g'(x)=\lim_{h\to 0} \frac{xf(x+h)-xf(x)+hf(x+h)}{h}$$

$$g'(x)=\lim_{h\to 0} (\frac{xf(x+h)-xf(x)}{h}+\frac{hf(x+h)}{h})$$

$$g'(x)=\lim_{h\to 0} (\frac{x[f(x+h)-f(x)]}{h}+f(x+h))$$

$$g'(x)=\lim_{h\to 0} (x\cdot\frac{f(x+h)-f(x)}{h}+f(x+h))$$

$$g'(x)=\lim_{h\to 0} (x\cdot\frac{f(x+h)-f(x)}{h})+\lim_{h\to 0}f(x+h)$$

$$g'(x)=\lim_{h\to 0}x \cdot\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}+\lim_{h\to 0}f(x+h)$$

$$g'(x)=\lim_{h\to 0}(x) \cdot f'(x)+\lim_{h\to 0}f(x+h)$$

$$g'(x)= x \cdot f'(x)+f(x+0)$$

$$g'(x)= x \cdot f'(x)+f(x)$$

I don't know if what I did is right. Let me know.

For the second one I don't even know how to start.

Any help will be very appreciated.

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    $\begingroup$ For the first one, continue with what you started. You can split your fraction up as $x\times \frac {f(x+h)-f(x)}h+f(x+h)$ so just take the limit, for the second, expand $f(x+h) $ according to the given rule. $\endgroup$
    – lulu
    Sep 14 '19 at 20:26
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    $\begingroup$ Using MSE to impress a girl ♡ $\endgroup$
    – Milan
    Sep 14 '19 at 20:48
  • $\begingroup$ @Milan LOL. I'm a lady, and she's family. I'm not trying to impress anyone. I'm just trying to help. $\endgroup$
    – gi2302
    Sep 14 '19 at 21:29
  • $\begingroup$ @lulu I edited the posting. Coud you please verify if I understood you well? Is that what you mean? I got what I would get if I used the product rule of derivative. So I'm thinking that I got it. But I want your confirmation please. Any clue on how to start with the second? $\endgroup$
    – gi2302
    Sep 14 '19 at 21:48
  • $\begingroup$ Yes, part $1$ looks good. I already gave you a hint for part $2$...just expand $f(x+h)$ according to the given rule. $\endgroup$
    – lulu
    Sep 14 '19 at 22:35
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Your first problem looks good. Remember that even though you weren't allowed to use the Product Rule to solve the problem, there's nothing stopping you from using it to verify the answer. ^_^

For the second part, let's rearrange the identity to $f(a+b)-f(a)=f(b)+a^2b+ab^2$ to make it more obvious what's going on. For part a: $$f(a+0)-f(a)=f(0)+a^2\cdot0+a\cdot0^2\\f(a)-f(a)=f(0)\\0=f(0)$$ For part b: $$f'(0)=\lim_{h\to0}\frac{f(0+h)-f(0)}h=\lim_{h\to0}\frac{f(h)-0^2\cdot h-0\cdot h^2}h\\=\lim_{h\to0}\frac{f(h)}h=1$$ And for part c: $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h=\lim_{h\to0}\frac{f(h)-x^2h-xh^2}h \\=\lim_{h\to0}\frac{f(h)}h+\lim_{h\to0}\frac{x^2h}h+\lim_{h\to0}\frac{xh^2}h\\ =\lim_{h\to0}\frac{f(h)}h+\lim_{h\to0}x^2+\lim_{h\to0}xh=1+x^2$$

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