# 6. Derivative of the Exponential Function

by M. Bourne

The derivative of *e ^{x} *is quite remarkable. The expression for the derivative is the same as the expression that we started with; that is,

*e*!

^{x}`(d(e^x))/(dx)=e^x`

**What does this mean? **It means the slope is the same as the function value (the *y*-value) for all points on the graph.

**Example: **Let's take the example when *x* = 2. At this point, the *y*-value is *e*^{2} ≈ 7.39.

Since the derivative of *e ^{x}* is

*e*, then the slope of the tangent line at

^{x}*x*= 2 is also

*e*

^{2}≈ 7.39.

We can see that it is true on the graph:

Let's now see if it is true at some other values of *x*.

When `x=0,` `y="slope"=1`.

When `x=3,` `y="slope"~~20.1`.

We can see that in each case, the **slope** of the curve `y=e^x` is the same as the **function value** at that point.

## Other Formulas for Derivatives of Exponential Functions

If *u* is a function of *x*, we can obtain the derivative of an expression in the form *e ^{u}*:

`(d(e^u))/(dx)=e^u(du)/(dx)`

If we have an exponential function with some base *b*, we have the following
derivative:

`(d(b^u))/(dx)=b^u ln b(du)/(dx)`

[These formulas are derived using first principles concepts. See the chapter on Exponential and Logarithmic Functions if you need a refresher on exponential functions before starting this section.]

### Example 1

Find the derivative
of *y* = 10^{3x}.

Answer

`y=10^(3x)`

So:

`(dy)/(dx)=10^(3x)(ln 10)((d(3x))/(dx))`

`=3 ln 10(10^(3x))`

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### Example 2

Find the derivative
of *y* = *e*^{x2}.

Answer

`y=e^(x^2)`

So:

`(dy)/(dx)=e^(x^2)((d(x^2))/(dx))`

`=2x(e^(x^2))`

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### Example 3

Find the derivative
of *y* = sin(*e*^{3x}).

Answer

`y=sin(e^(3x))`

Let `y=sin u`, where `u=e^(3x)`

`(dy)/(dx)=(dy)/(du)(du)/(dx)`

`=[cos u](d(e^(3x)))/(dx)`

`=[cos e^(3x)](3e^(3x))`

`=3e^(3x) cos e^(3x)`

### Example 4

Find the derivative
of *y* = *e*^{sin x}.

Answer

`y=e^(sin x)`

Let `u=sin x`, so `y=e^u`.

So:

`(dy)/(dx)=(dy)/(du)(du)/(dx)`

`=e^u (d(sin x))/(dx)`

`=e^(sin x) (cos x)`

`=cos x(e^(sin x))`

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### Example 5

Find the derivative of

`y=(ln 2x)/(e^(2x)+2`

Answer

`y=(ln 2x)/(e^(2x)+2`

We let *u* = ln 2*x* and *v* = *e*^{2x} + 2, and we'll use the derivative of a quotient formula,

`(dy)/(dx)=(v(du)/(dx)-u(dv)/(dx))/(v^2)`

We'll need to find the derivative of both *u* and *v* before using the formula.

Now, using the logarithm laws, we have:

u= ln 2x= ln 2 + ln

x.

So simply,

`(du)/(dx)=1/x`

And for* v* = *e*^{2x} + 2 we have:

`(dv)/(dx)=2e^(2x)`

So applying the quotient formula, we obtain:

`(dy)/(dx)=((e^(2x)+2)(d(ln 2x))/(dx)-ln 2x(d(e^(2x)+2))/(dx))/((e^(2x)+2)^2`

Using the derivatives we just found for *u* and *v *gives:

`(dy)/(dx)=((e^(2x)+2)(1/x)-ln 2x(2e^(2x)))/((e^(2x)+2)^2`

We tidy this up by multiplying top and bottom by *x.* Also, it's best to write the 2*e*^{2x} in front of the logarithm expression to reduce confusion. (It's not part of the log expression.)

So our derivative is:

`(dy)/(dx)=((e^(2x)+2)-2e^(2x)(x)(ln 2x))/(x(e^(2x)+2)^2)`

`(dy)/(dx)=((e^(2x)+2)-2x e^(2x)(ln 2x))/(x(e^(2x)+2)^2)`

## Exercises

1. Find the derivative of *y* = 10^{x2}.

Answer

Here, *b* = 10 and *u* =
*x*^{2}.

We have:

`(du)/(dx)=2x`

So

`(dy)/(dx)=10^(x^2)(ln 10)(2x)`

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2. Find the derivative of

`y=cos 2x(e^(x^2-1))`

Answer

This is a product.

Let *u* = cos 2*x* and `v=e^(x^2-1)`

Then `(du)/(dx)=-2 sin 2x`,

and `(dv)/(dx)=e^(x^2-1)(2x)`

So

`(dy)/(dx)=[cos 2x][(e^(x^2-1))(2x)]+` `[(e^(x^2-1))(-2 sin 2x)]`

`=(e^(x^2-1))[cos 2x(2x)-` `{:2 sin 2x]`

`=2e^(x^2-1)[x cos 2x-sin 2x]`

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3. Find the derivative of

`y=(2e^(x^2)+x^2)^3`

Answer

`y=(2e^(x^2)+x^2)^3`

Let

`u=2e^(x^2)+x^2`

then *y* =
*u*^{3}.

So

`(du)/(dx)=2(2x)e^(x^2)+2x`

`=4x e^(x^2)+2x`

and

`(dy)/(du)=3u^2`

`=3(2e^(x^2)+x^2)^2`

So

`(dy)/(dx)=(dy)/(du)(du)/(dx)`

`=3[2e^(x^2)+x^2]^2[4x e^(x^2)+2x]`

`=6x[2e^(x^2)+x^2]^2[2e^(x^2)+1]`

4. Show that

`y=e^(-x)sin x`

satisfies the (second order differential) equation

`(d^2y)/(dx^2)+2(dy)/(dx)+2y=0`

Answer

`y=e^(-x) sin x`

so

`(dy)/(dx)=e^(-x) cos x+sin x(-e^(-x))`

`=e^(-x)[cos x-sin x]`

and

`(d^2y)/(dx^2)=e^(-x)[-sin x-cos x]+` `[cos x-sin x][-e^(-x)]`

`=e^(-x)[-2 cos x]`

So

`"LHS"=(d^2y)/(dx^2)+2(dy)/(dx)+2y`

`=e^(-x)[-2 cos x]+` `2[e^(-x)(cos x-sin x)]+` `2[e^(-x) sin x]`

`=e^(-x)[-2 cos x+2 cos x -` `{:2 sin x+2 sin x]`

`=0`

`="RHS"`

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This last exercise is important in **electronics**, since the funtion `y=e^(-x)sin(x)` represents a **decaying** signal.

The graph of `y=e^(-x)sinx.`

We'll see more of these curves in Second Order Differential Equations, in the later calculus section.

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