# Using the derivative definition to prove these problems

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.

• 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.
– lulu
Sep 14 '19 at 20:26
• Using MSE to impress a girl ♡ Sep 14 '19 at 20:48
• @Milan LOL. I'm a lady, and she's family. I'm not trying to impress anyone. I'm just trying to help. Sep 14 '19 at 21:29
• @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? Sep 14 '19 at 21:48
• Yes, part $1$ looks good. I already gave you a hint for part $2$...just expand $f(x+h)$ according to the given rule.
– lulu
Sep 14 '19 at 22:35

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: $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h=\lim_{h\to0}\frac{f(h)-x^2h-xh^2}h \\=\lim_{h\to0}\frac{f(h)}h+\lim_{h\to0}\frac{x^2h}h+\lim_{h\to0}\frac{xh^2}h\\ =\lim_{h\to0}\frac{f(h)}h+\lim_{h\to0}x^2+\lim_{h\to0}xh=1+x^2$$