# Derivatives of Exponential Functions

Standard

HERE IS SOME BRIEF HELP FROM THE BOOK YOU CAN USE TO STUDY FOR NEXT CLASS! THIS IS REVIEW FROM PAGES 234-237. I HOPE THIS HELPS SOME!

First off, what is an exponential function?

A exponential  function is a function whose value is a constant raised to the power of the argument.

For example, f(x)=$a^{x}$, where a > 0 and a $\neq$ 1

FINDING THE DERIVATIVE:

We can find the derivative of exponential functions by using the definition of the derivative.

$\frac {d(a^x)}{dx}$$\overset {lim}{ h \rightarrow 0}$ $\frac {a^{x+h}-a^x}{h}$ (assuming a > 0)

=$\overset {lim}{ h \rightarrow 0}$ $\frac {a^{x}a^{h}-a^x}{h}$

=$\overset {lim}{ h \rightarrow 0}$ $\frac {a^{h}-1}{h}$

In this last step, since $a^x$ does not involve h, we were able to bring $a^x$ in front of the limit. The result says that the derivative of $a^x$  is $a^x$ times a constant that depends on a, namely $\overset {lim}{ h \rightarrow 0}$ $\frac {a^{h}-1}{h}$.

We could also find the derivative with the following short cuts:

derivative of $a^x$

$D_x [a^x]= (ln a) a^x$

This formula becomes particularly simple when we let a = e, because of the face that ln e = 1.

Derivative of $e^x$

$D_x[e^x] = e^x$

We now see why e is the best base to work with: it has the simplest derivative of all exponential functions. Even if we choose a different base, e appears in the derivative anyway through the ln a term. (Recall that ln a is tha logarithm of a to the base e.) In fact, of all the functions we have studied, $e^x$ is the simplest to differentiate, because its derivative is just itself.

Derivative of $a^{g(x)}$  and $e^{g(x)}$

$D_x[a^{g(x)}] = (ln a)a^{g(x)} g'(x)$

and

$D_x[e^{g(x)}] = e^{g(x)}g'(x)$

EXAMPLES:

(A) y= $e^{5x}$

Solution:  let g(x) = 5x, with g'(x) =5 then

$\frac {dy}{dx} = 5e^{5x}$

(B) $y = e^ {x} \sqrt {x}$

Solution: use the product rule

$\frac {dy}{dx} = e^x$ $\cdot$ $\frac {1}{2 \sqrt x}$  + $\sqrt {xe^x}$

= $\frac {e^x}{2 \sqrt x} + e^x \sqrt {x}$ $\cdot$ $\frac {2 \sqrt x}{2 \sqrt x}$

= $\frac{e^x}{2 \sqrt x} (1 + 2x)$

(C) f(x)= $\frac {100,000}{1+ 100e^{-0.3x}}$

Solution: use the quotient rule.

f(x) = $\frac {(1 + 100 e^{-0.3x})(0) - 100,000 (-30e^{-0.3x})}{(1 + 100 e^{-0.3x})}$

= $\frac {3,000,000e^{-0.3x}}{(1 + 100 e^{-0.3x})^ {2}}$

******Here is a joke that is actually in our textbook:*****

A deranged mathematician who frightened other inmates at an insane asylum by screaming at them, “I am going to differentiate you!” But one inmate remained calm and simply responded, ” I don’t care; I’m $e^x$.”