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.”

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