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
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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.
Determine the gradient to the curve at $x=0$x=0.
Evaluate $f\left(0\right)$f(0).
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!+. . .
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).
Round your answer to two decimal places.
question 4
Consider the curve with equation $y=e^x$y=ex.
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).
Hence find the equation of the tangent to the curve at point $Q$Q.
Does this tangent line pass through the point $R\left(-2,0\right)$R(−2,0)?
Yes
ANo
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 |
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| $=$= | $e^u\times a$eu×a |
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| $=$= | $ae^{ax}$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 |
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Example 2
Differentiate $y=\frac{e^{2x-5}-e^{-x}}{e^x}$y=e2x−5−e−xex.
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 | $=$= | $\frac{e^{2x-5}-e^{-x}}{e^x}$e2x−5−e−xex |
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| $=$= | $\frac{e^{2x-5}}{e^x}-\frac{e^{-x}}{e^x}$e2x−5ex−e−xex |
Split the fraction |
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| $=$= | $e^{2x-5-x}-e^{-x-x}$e2x−5−x−e−x−x |
Use index laws to write terms with a single base |
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| $=$= | $e^{x-5}-e^{-2x}$ex−5−e−2x |
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×ex−5−(−2)×e−2x |
| $=$= | $e^{x-5}+2e^{-2x}$ex−5+2e−2x |
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+e−5x.
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=ex−e3x+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 |
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| $=$= | $e^u\times f'\left(x\right)$eu×f′(x) |
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| $=$= | $f'\left(x\right)e^{f\left(x\right)}$f′(x)ef(x) |
Substitute $u=f\left(x\right)$u=f(x) |
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=e3x2−x−2.
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)=3x2−x−2. 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 |
| $=$= | $\left(6x-1\right)e^{3x^2-x-2}$(6x−1)e3x2−x−2 |
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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 |
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| $=$= | $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 |
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| $=$= | $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 |
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| $=$= | $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.