What is the derivative of $\ln(4^x)$? What is the derivative of $\ln(4^x)$  (which I believe is also equal to $x\ln4$)?
Is it $\dfrac{1}{x\ln4}$?
 A: Recall that for $a>0$, we have
$$\log(a^b) = b \log(a)$$
Also, note that
$$\dfrac{d}{dx}\left( cx\right) = c$$
I trust you can finish it from here.
A: The thing you are supposed to understand here is that constants don't do anything to derivatives -- they just factor out of the expression.
More precisely, we have for any function $f$ and a constant $c$ not depending on $x$ that
$$
\frac{d}{d x} c f(x) = c \frac{d}{d x} f(x)
$$
For example, since $\frac{d}{d x} (x^4) = 4x^3$, I can multiply by any constant and the derivative is just as straightforward:
$$
\frac{d}{d x} \left[ 2.33\pi^2 \cos (4) \right] x^4 
= \left[ 2.33\pi^2 \cos (4) \right] 4x^3
$$
This works since the $ 2.33\pi^2 \cos (4) $ is constant - it doesn't depend on $x$.
If you get this, you should be able to calculate derivatives like this easily.
A: if your question is $ln(4^x)$ then its answer will be $\ ln4$ and if your question is $(\ln4)^x$ then use this formula
$$\dfrac {d}{dx}a^x=a^x\cdot\ln{a}$$
so
$$\dfrac {d}{dx}{(\ln4)}^x=(\ln4)^x\cdot\ln{(\ln4)}$$
A: Hint : $\ln 4^x=y \implies x \ln 4=y$
Since $\ln4$ is a constant. The derivative is simply the constant, i.e $\ln 4$
