Why is the derivative of $3^x$ equal to $3^x \cdot \ln 3$

Question

Our teacher tells us to convert it this way $ 3^x = e^{\ln 3^x}= e^{x\cdot\ln 3}$ and then use the rule $e^u\cdot u'$ but I can't understand where $\ln$ comes from and how $\ln 3^x$ = $x\cdot \ln 3$.

Answer 1

Because by the definition of $\ln$ we have $\ln3^x=x\ln3$ and from here: $$\left(3^x\right)'=\left(e^{x\ln3}\right)'=e^{x\ln3}\cdot\ln3=3^x\ln3.$$ Actually, $\log_ab$ it's a number $c$ such that $a^c=b$. (Here, $a>0$, $b>0$ and $a\neq1$)

Id est, $a^{\log_ab}=b$.

For $a=e$ we obtain: $e^{\ln{b}}=b$ and $$3^x=\left(e^{\ln3}\right)^x=e^{x\ln3}.$$

Answer 2

And are you familiar with this basic property of logarithms? $$\log_{b}{x^y}=y\log_{b}{x}$$ You can bring the power out front.

How about this fact?

$$ a=e^{\ln{a}}, a >0 $$

Do you know what the derivative of an exponential function is?

$$ (a^x)'=a^x\ln{a} $$

This can be proven a number of ways. You can go back to the definition of the limit and prove it that way or you can use logarithmic differentiation:

$$ y=a^x\\ \ln{y}=\ln{a^x}\\ \ln{y}=x\ln{a}\\ (\ln{y})'=(x\ln{a})'\\ \frac{1}{y}y'=\ln{a}\\ y'=y\ln{a}\\ y'=a^x\ln{a} $$

Answer 3

Any term $x$ can be expressed as power of $e$:

$$x=e^{\ln x}$$