# Derivatives of Exponential and Logarithmic Functions

The final derivative type that we need to investigate is the derivatives of exponential and logarithmic functions.

First, let's refresh on the definition of exponential and logarithmic functions. An exponential function is a function in the form $f(x) = a^x$, where a is some constant in the real numbers. There is a special exponential function, which is $f(x) = e^x$. This constant, e, is known as Euler's number. It is a number that appears often in natural phenomena and has some special properties. The most interesting property to us is related to its derivative, which we will define now.

**Definition: **$\frac{dy}{dx}(e^x) = e^x$

This definition reveals what is so special about $e^x$. It is a function whose derivative is itself, which is something we haven't seen before. Let's prove that this is true.

$lim_{h \to 0} \frac{e^{x+h} - e^x}{h}$

$=lim_{h \to 0} \frac{e^xe^h - e^x}{h}$

$=lim_{h \to 0} \frac{e^x(e^h - 1)}{h}$

$=e^x * lim_{h \to 0} \frac{e^h - 1}{h}$

At this point, we are stuck since we have no way to evaluate $lim_{h \to 0} \frac{e^h - 1}{h}$. Let's see if we can find an answer to what it might be. Let's look at a graph of $\frac{e^h - 1}{h}$ and see what the functions behaviour is around h = 0.

As you can see, from the left and right the function appears to approach 1. This tells us that $e^x * lim_{h \to 0} \frac{e^h - 1}{h} = e^x * 1 = e^x$.

Therefore, $\frac{dy}{dx}(e^x) = e^x$

This result is important because we can prove the more general derivatives of exponentials and logarithms using it. Before we do this though, we need to find one more derivative related to the inverse of $e^x$.

For exponential functions, there exists an inverse operation, known as a logarithm. For some exponential, $a^x$, we have an inverse $log_a(a^x) = x$. For $e^x$, we define a special logarithm, called $ln(x)$. This is simply $log_e(x)$, we just give it a special name. We will next take a look at the derivative of $ln(x)$.

**Definition: **$\frac{dy}{dx}(ln(x)) = \frac{1}{x}$

To prove this, we can apply implicit differentiation.

$y = ln(x) \Longrightarrow e^y = x$

$\Longrightarrow \frac{dy}{dx}(e^y) = \frac{dy}{dx}(x)$

$\Longrightarrow e^y * \frac{dy}{dx} = 1$

$\Longrightarrow \frac{dy}{dx} = \frac{1}{e^y}$

Since we know that $e^y = x$, we can substitue this to get $\frac{dy}{dx}(ln(x)) = \frac{1}{x}$

Knowing both results, we can now take on the more general exponential and logarithmic functions.

**Definition: **$\frac{dy}{dx}(a^x) = a^x * ln(a)$

To prove this, we start by taking the ln of both sides of the equation. This gives us:

$ln(y) = ln(a^x)$

We can then apply power rules and implicit differentiation to get the result.

$ln(y) = ln(a^x)$

$\Longrightarrow ln(y) = x * ln(a)$

$\Longrightarrow \frac{dy}{dx}(ln(y)) = \frac{dy}{dx}(x * ln(a))$

$\Longrightarrow \frac{1}{y} * \frac{dy}{dx} = ln(a) + x * \frac{1}{x}$

$\Longrightarrow \frac{1}{y} * \frac{dy}{dx} = ln(a)$

$\Longrightarrow \frac{dy}{dx} = ln(a) * y$

Since we know that in this problem, $y = a^x$, we get: $\frac{dy}{dx} = ln(a) * a^x$

We can find the derivative of logarithmic functions in a similar way.

**Definition: **$\frac{dy}{dx}(log_a(x)) = \frac{1}{xln(a)}$

If $y = log_a(x)$, then $a^y = x$

From here we can apply implicit differentiation.

$\frac{dy}{dx}(a^y) = \frac{dy}{dx}(x)$

$\Longrightarrow a^yln(a)*\frac{dy}{dx} = 1$

$\Longrightarrow \frac{dy}{dx} = \frac{1}{a^yln(a)}$

Since we know that $a^y = x$, we can substitute it to get: $\frac{dy}{dx} = \frac{1}{xln(a)}$

This gives us a full set of rules to handle exponential functions, meaning we can now take the derivatives of most functions we encounter.