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

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 $$\ln 3^x$$ = $$x\cdot \ln 3$$.

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 $$3^x=\left(e^{\ln3}\right)^x=e^{x\ln3}.$$

• But why do we have to convert it to a natural number and why is there an ln in the exponent? Thats what I dont understand. Sorry if this is a stupid question – ythhtrg Mar 4 at 21:39
• @ythhtrg it's the definition of $\ln$. $\log_ab$ it's a number $c$ such that $a^c=b$. Id est, $a^{\log_ab}=b$. For $a=e$ we obtain: $e^{\ln{b}}=b$ and $3^x=\left(e^{\ln3}\right)^x=e^{x\ln3}.$ – Michael Rozenberg Mar 4 at 21:42

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.

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

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:

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

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

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