How to find the derivatives of powers of $f$ using the definition of derivative? have a homework problem that is just stumping me!
Basically I cannot use any derivative rules and I can only use the definition of derivative to solve these:
$$g(x)=f^2(x)$$
$$g(x)=f^3(x)$$
$$g(x)=\sqrt{f(x)}$$
$$g(x)=\sqrt[3]{f(x)}$$
I know how to use the definiton of derivative fairly well and I know what the answers should be, but I just can't get them to work out.
Any help would be much appreciated!
 A: For the first example, you secretly know the result $g'(x)=2f(x)f'(x)$, which may help as a guidance for a direct approach:
You have $$\begin{align}\frac{g(x+h)-g(x)}{h} &=\frac{f^2(x+h)-f^2(x)}{h}\\&=\frac{f^2(x+h)-f(x)f(x+h)+f(x)f(x+h)-f^2(x)}{h}\\&=(f(x+h)+f(x))\cdot \frac{f(x+h)-f(x)}{h}\end{align}$$ 
and see what happens as $h\to 0$, noting especialy that $f(x+h)\to f(x)$ holds because $f'(x)$ exists.
The same trick works for the second problem.
For the third problem, you can sort-of recycle the first result, as we find
$$\frac{f(x+h)-f(x)}{h} =(g(x+h)+g(x))\cdot\frac{g(x+h)-g(x)}{h}$$
from $f(x)=g^2(x)$. The reason for $g(x+h)\to g(x)$, however, is a different one (we cannot yet use the existence of $g'(x)$). Also be careful about the case $g(x)=0$.
Similarly, the fourth problem relates to the second.
A: Do you know the chain rule?  Assuming you are asked for $g'(x)$ in each case, for the third we can define $h(y)=y^3$, then $g(x)=h(f(x))=h'(f(x))f'(x)=3(f(x))^2f'(x)$  The others are similar
A: By Chain Rule, we know that the derivative of $[f(x)]^2$ should be $2f(x) \cdot f'(x)$. Indeed, observe that:
\begin{align*}
\frac{d}{dx} [f(x)]^2
&= \lim_{h \to 0} \frac{[f(x + h)]^2 - [f(x)]^2}{h} \\
&= \lim_{h \to 0} \frac{[f(x + h) + f(x)][f(x + h) - f(x)]}{h} \\
&= \lim_{h \to 0} \left[[f(x + h) + f(x)] \cdot \frac{f(x + h) - f(x)}{h}\right] \\
&= \left[\lim_{h \to 0} [f(x + h) + f(x)]\right] \cdot \left[\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\right] \\
&= [f(x + 0) + f(x)] \cdot f'(x) \\
&= 2f(x) \cdot f'(x)
\end{align*}
as desired. Try something similar for the other problems.
