Q: Using Definition of Derivative To Solve for Derivative I am looking for any assistance with knowing where to start for the following two problems:


*

*Consider the following equation:
$$ f'(a) = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$$


Let $ f(x) = \sqrt[3]{x} $
If a ≠ 0, use the above formula to find $ f'(a) $


*Suppose f is a function which satisfies the properties:
$$ f(x+y) = f(x) + f(y) + x^2y + xy^2 $$
For all real numbers x and y, and
$$ \lim\limits_{x \to 0} \frac{f(x)}{x} = 1 $$
Find $ f'(x) $


A: we have $$\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}=\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}\frac{\sqrt[3]{x+h}^2+\sqrt[3]{x+h}\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt[3]{x+h}^2+\sqrt[3]{x+h}\sqrt[3]{x}+\sqrt[3]{x^2}}=...$$
can you proceed?

A: For number 1 it is just a simple proof of power rule. You can derive it generally for $x^n$ by the binomial theorem. For number 2 we have $f'(x)=\displaystyle\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$ by what the question tells us we know that this comes out to $\displaystyle\lim\limits_{h\to0}\frac{f(x)+f(h)+x^2h+xh^2-f(x)}{h}$ $=\displaystyle\lim\limits_{h\to0}\frac{f(h)}{h}+x^2+xh=1+x^2$

A: *

*For every $a, h\ne 0$ we have
\begin{eqnarray}
\dfrac{f(a+h)-f(a)}{h}&=&\dfrac{\sqrt[3]{a+h}-\sqrt[3]{a}}{h}=\dfrac{(\sqrt[3]{a+h}-\sqrt[3]{a})[(\sqrt[3]{a+h})^2+\sqrt[3]{a}\sqrt[3]{a+h}+(\sqrt[3]{a})^2]}{h[(\sqrt[3]{a+h})^2+\sqrt[3]{a}\sqrt[3]{a+h}+(\sqrt[3]{a})^2]}\\
&=&\dfrac{(\sqrt[3]{a+h})^3-(\sqrt[3]{a})^3}{h[(\sqrt[3]{a+h})^2+\sqrt[3]{a}\sqrt[3]{a+h}+(\sqrt[3]{a})^2]}\\
&=&\dfrac{a+h-a}{h[(\sqrt[3]{a+h})^2+\sqrt[3]{a}\sqrt[3]{a+h}+(\sqrt[3]{a})^2]}\\
&=&\dfrac{h}{h[(\sqrt[3]{a+h})^2+\sqrt[3]{a}\sqrt[3]{a+h}+(\sqrt[3]{a})^2]}\\
&=&\dfrac{1}{(\sqrt[3]{a+h})^2+\sqrt[3]{a}\sqrt[3]{a+h}+(\sqrt[3]{a})^2}\\
\end{eqnarray}
Taking the limit as $h$ goes to zero, we get
\begin{eqnarray}
\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}=\lim_{h\to0}\dfrac{1}{(\sqrt[3]{a+h})^2+\sqrt[3]{a}\sqrt[3]{a+h}+(\sqrt[3]{a})^2}=\dfrac{1}{3\sqrt[3]{a^2}}
\end{eqnarray}
Hence
$$
f'(a)=\dfrac{1}{3\sqrt[3]{a^2}}
$$

*For every $h\ne 0$ we have
$$
\dfrac{f(x+h)-f(x)}{h}=\dfrac{f(x)+f(h)+x^2h+xh^2-f(x)}{h}=\dfrac{f(h)}{h}+x^2+xh
$$
Taking the limit, we get
$$
\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}=
\lim_{h\to0}\left[\dfrac{f(h)}{h}+x^2+xh\right]=1+x^2
$$
Hence
$$
f'(x)=1+x^2
$$

