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2.05 Differentiation and exponential functions

Lesson

Derivative of $y=e^x$y=ex

In the investigation it was established that the function $y=e^x$y=ex has the remarkable property of being its own derivative.

That is for $y=e^x$y=ex, $\frac{dy}{dx}=e^x$dydx=ex.

What does a function being its own derivative actually mean? As well as simplifying many calculations in calculus this gives the graph the following properties:

  • This means the value of the function at any point on the graph is equal to the gradient of the tangent at that point
  • Equivalently the value of the function and the instantaneous rate of change are the same for any given value of $x$x

The value of the function at $x=1$x=1 is $e$e.

The gradient of the tangent at this point is $e$e.

Practice questions

question 1

$f\left(x\right)=e^x$f(x)=ex and its tangent line at $x=0$x=0 are graphed on the coordinate axes.

Loading Graph...

  1. Determine the gradient to the curve at $x=0$x=0.

  2. Evaluate $f\left(0\right)$f(0).

  3. Which of the following is true?

    $f\left(0\right)\ne f'\left(0\right)$f(0)f(0)

    A

    $f\left(0\right)=f'\left(0\right)$f(0)=f(0)

    B

question 2

The value of $e^x$ex, where $e$e is the natural base, can be given by the expansion below:

$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\text{. . .}$ex=1+x1!+x22!+x33!+x44!+x55!+. . .

  1. Find $\frac{d}{dx}\left(e^x\right)$ddx(ex) by filling in the gaps below.

    $\frac{d}{dx}\left(e^x\right)$ddx(ex) $=$= $1+\frac{2\editable{}}{2!}+\frac{\editable{}x^2}{3!}+\frac{\editable{}}{4!}+\frac{5\editable{}}{\editable{}}$1+22!+x23!+4!+5+$...$...
      $=$= $1+\frac{\editable{}}{1!}+\frac{x^2}{2!}+\frac{x^3}{\editable{}}+\frac{\editable{}}{4!}$1+1!+x22!+x3+4!+$...$...

    From this, we can deduce that the derivative of $y=e^x$y=ex is $\frac{dy}{dx}=\editable{}$dydx=

question 3

Find the gradient to the curve $y=e^x$y=ex at the point $\left(-1,\frac{1}{e}\right)$(1,1e).

  1. Round your answer to two decimal places.

question 4

Consider the curve with equation $y=e^x$y=ex.

  1. Determine the value of the gradient $m$m of the tangent to $y=e^x$y=ex at the point $Q\left(-1,\frac{1}{e}\right)$Q(1,1e).

  2. Hence find the equation of the tangent to the curve at point $Q$Q.

  3. Does this tangent line pass through the point $R\left(-2,0\right)$R(2,0)?

    Yes

    A

    No

    B

 

Derivative of $y=e^{ax}$y=eax

We now use the chain rule to find the derivative of the more general function given by $y=e^{ax}$y=eax:

 

Setting $u=ax$u=ax, we have that $y=e^u$y=eu. We then observe that:

$\frac{du}{dx}=a$dudx=a and $\frac{dy}{du}=e^u$dydu=eu

Putting this together, we have that:

$\frac{dy}{dx}$dydx $=$= $\frac{dy}{du}\times\frac{du}{dx}$dydu×dudx

 

  $=$= $e^u\times a$eu×a

 

  $=$= $ae^{ax}$aeax

Substitute $u=ax$u=ax

 

Recall we can also think of the chain rule as the derivative of the inside function multiplied by the derivative of the outside function.

Worked examples

Example 1

Differentiate $y=e^{5x}$y=e5x.

Think: We can see the function is a composite of $g\left(x\right)=5x$g(x)=5x as a function within $f\left(x\right)=e^x$f(x)=ex therefore we need to use the chain rule to differentiate $y=e^{5x}$y=e5x.

Do:

$\frac{dy}{dx}$dydx $=$= $5\times e^{5x}$5×e5x

Multiply the derivative of the inside function by the outside function

  $=$= $5e^{5x}$5e5x

 

Example 2

Differentiate $y=\frac{e^{2x-5}-e^{-x}}{e^x}$y=e2x5exex.

Think: We could use the quotient rule here but we could also split the fraction into parts. Here, with the single term of $e^x$ex in the numerator the split fractions can be simplified.

Do:

$y$y $=$= $\frac{e^{2x-5}-e^{-x}}{e^x}$e2x5exex

 

  $=$= $\frac{e^{2x-5}}{e^x}-\frac{e^{-x}}{e^x}$e2x5exexex

Split the fraction

  $=$= $e^{2x-5-x}-e^{-x-x}$e2x5xexx

Use index laws to write terms with a single base

  $=$= $e^{x-5}-e^{-2x}$ex5e2x

Simplify

 

Hence, differentiating term by and using the chain rule we obtain:

$\frac{dy}{dx}$dydx $=$= $1\times e^{x-5}-\left(-2\right)\times e^{-2x}$1×ex5(2)×e2x
  $=$= $e^{x-5}+2e^{-2x}$ex5+2e2x

 

Differentiating exponential functions

If $y=e^x$y=ex, then $\frac{dy}{dx}=e^x$dydx=ex.

If $y=e^{ax}$y=eax, then $\frac{dy}{dx}=ae^{ax}$dydx=aeax.

Practice questions

Question 5

Find the derivative of $y=e^{2x}+e^9+e^{-5x}$y=e2x+e9+e5x.

Question 6

Find the equation of the tangent to the curve $f\left(x\right)=2e^x$f(x)=2ex at the point where it crosses the $y$y-axis.

Express the equation in the form $y=mx+c$y=mx+c.

Question 7

Find the derivative of $y=\frac{e^x-e^{3x}+1}{e^x}$y=exe3x+1ex.

Differentiating $y=e^{f\left(x\right)}$y=ef(x)

Differentiating $y=e^{ax}$y=eax is a special case of the function $y=e^{f\left(x\right)}$y=ef(x), let's look more generally at differentiating functions of this form.

Consider functions of the form:

$y$y $=$= $e^{f\left(x\right)}$ef(x)
 

Setting $u=f\left(x\right)$u=f(x), we have $y=e^u$y=eu. We then observe that:

$\frac{du}{dx}=f'\left(x\right)$dudx=f(x) and $\frac{dy}{du}=e^u$dydu=eu

Using the chain rule:

$\frac{dy}{dx}$dydx $=$= $\frac{dy}{du}\times\frac{du}{dx}$dydu×dudx

 

  $=$= $e^u\times f'\left(x\right)$eu×f(x)

 

  $=$= $f'\left(x\right)e^{f\left(x\right)}$f(x)ef(x)

Substitute $u=f\left(x\right)$u=f(x)

 

Exponential functions and the chain rule

If $y=e^{f\left(x\right)}$y=ef(x), then $\frac{dy}{dx}=f'\left(x\right)e^{f\left(x\right)}$dydx=f(x)ef(x).

We can be expected to differentiate more complex exponential functions with the chain rule, the product rule, the quotient rule or combinations of these. Let's look at some examples.

Worked examples

Example 3

Differentiate the function $y=e^{3x^2-x-2}$y=e3x2x2.

Think: We can see the function is of the form $y=e^{f\left(x\right)}$y=ef(x), with $f\left(x\right)=3x^2-x-2$f(x)=3x2x2. Therefore we need to use the chain rule to differentiate.

Do: Using the chain rule:

$\frac{dy}{dx}$dydx $=$= $f'\left(x\right)\times e^{f\left(x\right)}$f(x)×ef(x)

Multiply the derivative of the inside function by the outside function

  $=$= $\left(6x-1\right)e^{3x^2-x-2}$(6x1)e3x2x2

 

Example 4

Differentiate the function $y=e^{4x}\left(2x+5\right)^7$y=e4x(2x+5)7, giving your answer in factorised form.

Think: Here we have the product of two functions, so we will use the product rule together with the chain rule.

Do:

Let $u=e^{4x}$u=e4x then $u'=4e^{4x}$u=4e4x by chain rule
and $v=\left(2x+5\right)^7$v=(2x+5)7 then

$v'=7\left(2x+5\right)^6\times2$v=7(2x+5)6×2 by chain rule

$\therefore v'=14\left(2x+5\right)^6$v=14(2x+5)6

$y'$y $=$= $uv'+vu'$uv+vu

Write out the product rule

  $=$= $e^{4x}\times14\left(2x+5\right)^6+\left(2x+5\right)^7\times4e^{4x}$e4x×14(2x+5)6+(2x+5)7×4e4x

Make the appropriate substitutions

  $=$= $14e^{4x}\left(2x+5\right)^6+4e^{4x}\left(2x+5\right)^7$14e4x(2x+5)6+4e4x(2x+5)7

To give the answer in factorised form look for common factors

  $=$= $2e^{4x}\left(2x+5\right)^6\left(7+2\left(2x+5\right)\right)$2e4x(2x+5)6(7+2(2x+5))

Take out common factors

  $=$= $2e^{4x}\left(2x+5\right)^6\left(4x+17\right)$2e4x(2x+5)6(4x+17)

Simplify


Practice questions

Question 8

Find the derivative of $y=\left(e^{2x}+5x\right)^{\frac{3}{4}}$y=(e2x+5x)34.

Question 9

Find the derivative of $y=x^5e^{4x}$y=x5e4x. Express the derivative in factorised form.

(Note: You may let $u=x^5$u=x5)

Question 10

Differentiate $y=\frac{e^{6x}}{1+e^x}$y=e6x1+ex.

Outcomes

ACMMM098

estimate the limit of [a^h−1]/h as h→0 using technology, for various values of a>0

ACMMM099

recognise that e is the unique number a for which the limit (in ACMMM098) is 1

ACMMM100

establish and use the formula d/dx(e^x)=e^x

ACMMM106

apply the product, quotient and chain rule to differentiate functions such as xe^x, tan⁡x,1/x^n, x sin⁡x, e^(−x)sin⁡x and f(ax+b)

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