Using the definition of a derivative, solve $\lim_{x \to 0} \frac{(2+h)^{3+h} - 8}{h}$ I can't figure out where to even start, I have looked up the answer on Desmos but it uses L'Hopitals rule which i haven't learned yet.
$$\lim_{h \to 0} \frac{(2+h)^{3+h} - 8}{h}$$
I see that i can use the log rules to rewrite it as
$$\lim_{h \to 0} \frac{e^{(3+h)\ln(2+h)} - 8}{h}$$
but that only confuses me more.
 A: Per Levent's comment, supposed that you're given $f(x) = x^{x+1}$.  By definition of the derivative:
$$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{(x+h)^{x+1+h} - x^{x+1}}{h}$$
Now, plug in $x=2$.
$$f'(2) = \lim_{h \rightarrow 0} \frac{(2+h)^{3+h} - 8}{h}$$
Hey, that's conveniently familiar!  Now your question can be rephrased as “Find $f'(2)$.”
Now, what's the derivative of $y = f(x) = x^{x+1}$?  The equation may be easier to work with if you take logarithms.
$$ \ln y = (x + 1) \ln x$$
You can then use implicit differentiation to get:
$$\frac{1}{y} \frac{dy}{dx} = \ln x + \frac{x+1}{x} = \ln x + 1 + \frac{1}{x}$$
$$\frac{dy}{dx} = y(\ln x + 1 + \frac{1}{x})$$
And plugging in $x=2$ and $y=8$ gives:
$$f'(2) = 8(\ln 2 + 1 + \frac{1}{2}) = 8 \ln 2 + 12$$
A: I follow up on your suggestion to use a logarithm. Define
$$ \phi(h)= e^{(3+h)\ln(2+h)}, \text{ noting that } \phi(0)= 8.$$
On the one hand, $$\phi'(0) \;=\; \lim_{h \to 0}\frac{\phi(h)-\phi(0)}{h-0},$$ which is exactly your target.
But applying the basic rules for derivatives of complicated expressions:
$$ \phi'(h)= \phi(h)\left[\ln(2+h) + (3+h)\frac1{2+h}\right],$$
$$ \phi'(0)= 12+8\ln 2.$$
