Let's say I have a simple problem like this:
$$
\frac{d}{dx}[2^x]
$$
I would apply the exponent rule and work it like this:
$$
\frac{d}{dx}[2^x]
$$
$$
=x\cdot 2^{x-1}
$$
**I don't know how to simplify that further so I will leave it as is. If anyone knows, just comment.
But the real answer turns out to be $\ln(2)\cdot 2x$. How is this worked out and how can I apply this rule to complex problems like:
$$
\frac{d}{dx}[(2x+4)^{x+1}]
$$
I want to know the following:
- How to use this $\ln(x)$ rule?
- When to use this rule?
- Why you can't use the standard exponent rule?
Best Answer
I think that you are confusing the rules. The rule you are misquoting is $$\frac{d}{dx} x^a =ax^{a-1}$$ (Note that $x$ is not in the exponent). To calculate the derivative of $a^x$ we will use the special property of $e$. More precisely, we have: $$\begin{align}\frac{d}{dx}a^x&=\frac{d}{dx} e^{x\ln a}\\[1ex]&=e^{x\ln a}\left(\dfrac{d}{dx}x\ln a\right)\\[1ex]&=e^{x\ln a}\ln a\\[1ex]&=a^x \ln a\end{align}$$
So for the more complex example, we have: $$\begin{align} \frac{d}{dx}[(2x+4)^{x+1}] &= \frac{d}{dx} e^{(x+1)\ln(2x+4)}\\[1ex]&= \left(\frac{d}{dx}((x+1)\ln(2x+4))\right)e^{(x+1)\ln(2x+4)}\\[1ex]&=\left(\ln(2x+4) +\frac{x+1}{2x+4}\right)e^{(x+1)\ln(2x+4)}\\[1ex]&=\left(\ln(2x+4)+\frac{1}{2}\frac{x+1}{x+2}\right)(2x+4)^{x+1} \end{align}$$
If you want a more general rule we can write: $$\begin{align}\frac{d}{dx} f(x)^{g(x)}&=\frac{d}{dx}e^{g(x)\ln(f(x))}\\[1ex]&=\left(g'(x)\ln(f(x))+g(x)\frac{1}{f(x)}\right)e^{g(x)\ln(f(x))}\\[1ex]&=\left(g'(x)\ln(f(x))+g(x)\frac{1}{f(x)}\right) f(x)^{g(x)}\end{align}$$
Hope that helps.