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Derivatives of exponential and logarithmic functions calculated.

### The Derivative of the Natural Logarithm

We do not yet have a shortcut formula for the derivative of the natural logarithm, so let’s start from the definition. Set $f(x) = \ln x$.

That is, The limit inside the logarithm is a bit beyond what we can deal with right now, so unless we can come up with a different strategy, we’re stuck.

What do we know about this logarithm? We know that the natural logarithm function $f(x) = \ln x$ is the inverse of the exponential function $e^x$ That is, Since we’re trying to find the derivative of $\ln x$, that means we’re trying to find $y'$. Rather than working with the logarithmic version of $y = \ln x$, let’s try to work with its exponential version $e^y = x$. We’ll start by taking the derivative of both sides The right-hand side is easy, but what about the left-hand side? If we think of $y$ as just a function of $x$, then the left-hand side is the exponential $e^x$ with the $x$ replaced by a function. It’s a Chain Rule problem, when we think of $e^y$ as having an outside function $g(x) = e^x$ and an inside function $f(x) = y$. By Chain Rule, $\ddx g(f(x)) = g'(f(x)) f'(x) = e^y y'$. Let’s put all this together.

We notice once again that $e^y = x$, so $\dfrac {1}{e^y} = \dfrac {1}{x}$. This gives our derivative formula.

Compute $\displaystyle \ddx e^{3x} \ln \left ( 4x^2+3\right )$.
$\displaystyle e^{3x} \dfrac {1}{4x^2+3}$ $\displaystyle 3e^{3x} \dfrac {8x}{4x^2+3}$ $\displaystyle 3e^{3x}\ln (4x^2+3) + e^{3x} \dfrac {8x}{4x^2+3}$ $\displaystyle 3e^{3x} + \dfrac {8x}{4x^2+3}$

### The Derivative of Exponentials and Logarithms with Other Bases

We have found derivative formulas for the natural exponential function $e^x$ and the natural logarithm function $\ln x$, but we have not yet explored other bases. That will be our focus for the rest of the section.

For exponentials, we remember that any number $b > 0$ can be written in the form $e^x$ for some specific value of $x$. To determine the $x$, we solve the equation $b = e^x$ so $x = \ln b$. That is, $b = e^{\ln b}$.

To find $\ddx b^x$ we are finding $\ddx e^{x \ln b}$, which we know by Chain Rule.

Rewriting $e^{x \ln b}$ as $b^x$ we find our derivative formula.

To deal with logarithms of other bases, we rely on the change of base formula:

This formula allows us to replace a logarithm with one base with a logarithm with whatever base we want. There is one base that we like more than the rest, base $e$. This means $\displaystyle \log _b x = \dfrac { \ln x}{ \ln b}$.

This gives our derivative formula.

Compute $\displaystyle \ddx 6^{x^2} \log _5\left (x+3\right )$.
$\displaystyle 6^{2x} \dfrac {1}{(x+3) \ln 5}$ $\displaystyle 2x \cdot 6^{x^2}\ln (6)\log _5(x+3) + 6^{x^2} \dfrac {1}{(x+3) \ln 5}$ $\displaystyle x^2 \cdot 6^{x^2-1} \log _5(x+3) + 6^{x^2} \dfrac {1}{(x+3) \ln 5}$ $\displaystyle 2x \cdot 6^{x^2}\ln (6)\log _5(x+3) + 6^{x^2} \dfrac {5}{x+3}$