We have established the result that if $y=e^x$y=ex, then the derivative is given by:
$\frac{dy}{dx}=e^x$dydx=ex
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 |
Worked examples
Example 1
Differentiate $y=e^{5x}$y=e5x.
Think: The derivative of $y=e^x$y=ex is $e^x$ex. Since $5x$5x is a function within $y=e^x$y=ex therefore we can use the chain rule to differentiate $y=e^{5x}$y=e5x.
Do:
| $\frac{dy}{dx}$dydx | $=$= | $e^{5x}\times5$e5x×5 |
| $=$= | $5e^{5x}$5e5x |
Example 2
Differentiate $y=\frac{e^{2x}-e^{-x}}{e^x}$y=e2x−e−xex.
| $y$y | $=$= | $\frac{e^{2x}-e^{-x}}{e^x}$e2x−e−xex |
| $=$= | $\frac{e^{2x}}{e^x}-\frac{e^{-x}}{e^x}$e2xex−e−xex | |
| $=$= | $e^x-e^{-2x}$ex−e−2x | |
| $\therefore$∴$\frac{dy}{dx}$dydx | $=$= | $e^x+2e^{-2x}$ex+2e−2x |
Example 3
Differentiate $y=3e^{4x}$y=3e4x.
Think: The derivative of $y=3e^x$y=3ex is $3e^x$3ex. Since $4x$4x is a function within $y=3e^x$y=3ex therefore we can use the chain rule to differentiate $y=3e^{4x}$y=3e4x.
Do:
| $\frac{dy}{dx}$dydx | $=$= | $3e^{4x}\times4$3e4x×4 |
| $=$= | $12e^{4x}$12e4x |
Tangents and normals of exponential functions
The same rules we use for finding the equation of tangents and normals to a function also apply for exponential functions.
To find the equation of the tangent to the curve given by $y=e^{2x}$y=e2x at the point where $x=3$x=3, we first note that at $x=3$x=3:
| $y$y | $=$= | $e^{2\left(3\right)}$e2(3) |
| $y$y | $=$= | $e^6$e6 |
The derivative of $y=e^{2x}$y=e2x is:
| $\frac{dy}{dx}$dydx | $=$= | $2e^{2x}$2e2x |
And so at $x=3$x=3:
| $\frac{dy}{dx}$dydx | $=$= | $2e^6$2e6 |
Using the point-gradient form, the equation of the tangent becomes:
| $y-y_1$y−y1 | $=$= | $m\left(x-x_1\right)$m(x−x1) |
| $y-e^6$y−e6 | $=$= | $2e^6\left(x-3\right)$2e6(x−3) |
| $y-e^6$y−e6 | $=$= | $2e^6x-6e^6$2e6x−6e6 |
| $y$y | $=$= | $2e^6x-5e^6$2e6x−5e6 |
$\frac{d}{dx}\left(e^x\right)=e^x$ddx(ex)=ex
$\frac{d}{dx}\left(e^{ax}\right)=ae^{ax}$ddx(eax)=aeax
Practice questions
Question 1
Consider the function $y=e^{ax}$y=eax, where $a$a is a constant.
Let $u=ax$u=ax.
Rewrite the function $y$y in terms of $u$u.
Determine $\frac{du}{dx}$dudx.
Hence determine $\frac{dy}{dx}$dydx.
Consider the function $f\left(x\right)=e^{2x}$f(x)=e2x.
State $f'\left(x\right)$f′(x).
Consider the function $f\left(x\right)=e^{2x}$f(x)=e2x.
State $f'$f′$\left(-3\right)$(−3). Give your answer in exact form.
Question 2
Find the derivative of $y=e^{2x}+e^9+e^{-5x}$y=e2x+e9+e−5x.
Question 3
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 4
Find the derivative of $y=\frac{e^x-e^{3x}+1}{e^x}$y=ex−e3x+1ex.
Differentiating exponentials with other bases
To differentiate an exponential with a base other than $e$e we must rewrite the exponential in terms of $e$e.
Worked example
example 4
Differentiate $y=2^x$y=2x.
Think: We need to rewrite the function in terms of $e$e. To balance the equation we also need to use the natural log function $\ln$ln.
Do:
| $y$y | $=$= | $e^{\ln2^x}$eln2x |
We must introduce both e and the natural log to balance the equation. |
| $y'$y′ | $=$= | $e^{\ln2^x}\times\frac{d}{dx}\left(\ln2^x\right)$eln2x×ddx(ln2x) |
Using the chain rule to differentiate. |
| $=$= | $e^{\ln2^x}\times\frac{d}{dx}\left(x\ln2\right)$eln2x×ddx(xln2) |
Use log laws to move the power to the front of the log. |
|
| $=$= | $e^{\ln2^x}\times\ln2$eln2x×ln2 |
|
|
| $=$= | $2^x\ln2$2xln2 |
Now we can simplify the first part. |