We saw in the year 11 course that we can differentiate the exponential function as follows:
| $\frac{d}{dx}e^x$ddxex | $=$= | $e^x$ex |
| $\frac{d}{dx}e^{ax}$ddxeax | $=$= | $ae^{ax}$aeax |
We will now look at how to differentiate more general forms of the exponential function.
Differentiating $y=e^{f\left(x\right)}$y=ef(x)
Consider functions of the form:
| $y$y | $=$= | $e^{f\left(x\right)}$ef(x) |
Letting $u=f\left(x\right)$u=f(x), we have:
| $y$y | $=$= | $e^u$eu |
| $\frac{dy}{du}$dydu | $=$= | $e^u$eu |
And:
| $\frac{du}{dx}$dudx | $=$= | $f'\left(x\right)$f′(x) |
Now using the chain rule, we find that:
| $\frac{dy}{dx}$dydx | $=$= | $\frac{dy}{du}\times\frac{du}{dx}$dydu×dudx |
| $\frac{dy}{dx}$dydx | $=$= | $e^u\times u'$eu×u′ |
| $\frac{dy}{dx}$dydx | $=$= | $f'\left(x\right)e^{f\left(x\right)}$f′(x)ef(x) |
| $\frac{d}{dx}e^{f\left(x\right)}$ddxef(x) | $=$= | $f'\left(x\right)e^{f\left(x\right)}$f′(x)ef(x) |
Worked examples
Example 1
Differentiate the function $y=e^{3x^2-x-2}$y=e3x2−x−2.
Solution: Using the rule that we developed above, we have that:
| $\frac{dy}{dx}$dydx | $=$= | $\left(6x-1\right)e^{3x^2-x-2}$(6x−1)e3x2−x−2 |
Example 2
Differentiate the function $y=\frac{e^{2x}-e^{-x}}{e^x}$y=e2x−e−xex.
Think: This function has exponentials of the same base in its numerator and denominator. So it will be easier to differentiate if we first rewrite the function by splitting the fraction and simplifying the exponentials.
Do: Rewriting the function first:
| $y$y | $=$= | $\frac{e^{2x}-e^{-x}}{e^x}$e2x−e−xex |
| $y$y | $=$= | $\frac{e^{2x}}{e^x}-\frac{e^{-x}}{e^x}$e2xex−e−xex |
| $y$y | $=$= | $e^x-e^{-2x}$ex−e−2x |
Now we can more easily differentiate:
| $\frac{dy}{dx}$dydx | $=$= | $e^x+2e^{-2x}$ex+2e−2x |
Practice questions
Question 1
Differentiate $y=\left(2x-e^{3x}\right)^7$y=(2x−e3x)7.
Question 2
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 3
Differentiate $y=\frac{e^{6x}}{1+e^x}$y=e6x1+ex.
Differentiating $y=a^x$y=ax
In calculus, there are functions that first need to be manipulated into another form in order for them to be differentiated. The function $y=a^x$y=ax fits this criterion.
We know that the exponential function is a special case. When $y=e^x$y=ex is differentiated it gives $y'=e^x$y′=ex. That is, when the base is $e\approx2.7182818...$e≈2.7182818...:
$y=\frac{dy}{dx}=e^x$y=dydx=ex
Consider the applet below which shows the function $y=a^x$y=ax and its derivative (dotted line). Notice that the derivative of any exponential function, where $a\ne e$a≠e, is another similar exponential function.
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From the applet, we could say that the derivative of $y=a^x$y=ax, is the original function multiplied by some constant. That is:
$\frac{dy}{dx}=ka^x$dydx=kax, where $k$k is a constant.
We will use logarithm laws to determine this constant and find the derivative of $y=a^x$y=ax for the case when $a\ne e$a≠e.
| $y$y | $=$= | $a^x$ax |
Taking the natural log of both sides gives:
| $\ln y$lny | $=$= | $\ln a^x$lnax |
Using our log laws gives:
| $\ln y$lny | $=$= | $x\ln a$xlna |
Reversing the process, we get:
| $e^{\ln y}$elny | $=$= | $e^{x\ln a}$exlna |
Which can be simplified to:
| $y$y | $=$= | $e^{x\ln a}$exlna |
Taking the derivative of both sides using $\frac{d}{dx}e^{f(x)}=f'\left(x\right)e^{f\left(x\right)}$ddxef(x)=f′(x)ef(x):
| $\frac{dy}{dx}$dydx | $=$= | $\ln ae^{x\ln a}$lnaexlna |
But, remember $y=e^{x\ln a}=a^x$y=exlna=ax, therefore:
| $\frac{dy}{dx}$dydx | $=$= | $\ln a\ a^x$lna ax |
Therefore the constant term is the natural log of the base number. This agrees with what we know for the exponential function $y=e^x$y=ex, as $\ln e=1$lne=1.
| $\frac{d}{dx}a^x$ddxax | $=$= | $\ln a\ a^x$lna ax |
Worked example
Example 3
Differentiate the function $y=2^x$y=2x.
Think: This is an exponential function with $a\ne e$a≠e. Therefore we will use $\frac{d}{dx}a^x=\ln a\ a^x$ddxax=lna ax.
Do:
| $y$y | $=$= | $2^x$2x |
| $y'$y′ | $=$= | $\ln2\ 2^x$ln2 2x |
Practice question
Question 4
Consider the function $y=7^x$y=7x.
Using the fact that $e^{\ln a}=a$elna=a, rewrite the function in terms of natural base $e$e.
Enter each line of work as an equation.
Hence determine $y'$y′. Express the derivative in terms of the base $7$7.
You may use the substitution $u=\left(\ln7\right)x$u=(ln7)x.
Hence determine the exact gradient at $x=1$x=1.
Differentiating $y=a^{f\left(x\right)}$y=af(x)
Repeating the processes shown above we can find derivatives of functions in the form $y=a^{f\left(x\right)}$y=af(x).
Again we will use logarithm laws to determine the derivative of $y=a^{f\left(x\right)}$y=af(x) for the case when $a\ne e$a≠e.
| $y$y | $=$= | $a^{f\left(x\right)}$af(x) |
Taking the natural log of both sides gives:
| $\ln y$lny | $=$= | $\ln a^{f\left(x\right)}$lnaf(x) |
Using our log laws gives:
| $\ln y$lny | $=$= | $f\left(x\right)\ln a$f(x)lna |
Reversing the process, we get:
| $e^{\ln y}$elny | $=$= | $e^{f\left(x\right)\ln a}$ef(x)lna |
Which can be simplified to:
| $y$y | $=$= | $e^{f\left(x\right)\ln a}$ef(x)lna |
Taking the derivative of both sides using $\frac{d}{dx}e^{f\left(x\right)}=f'\left(x\right)e^{f\left(x\right)}$ddxef(x)=f′(x)ef(x):
| $\frac{dy}{dx}$dydx | $=$= | $\ln a\ f'(x)\ e^{f\left(x\right)\ln a}$lna f′(x) ef(x)lna |
But, remember $y=e^{f\left(x\right)\ln a}=a^{f\left(x\right)}$y=ef(x)lna=af(x), therefore:
| $\frac{dy}{dx}$dydx | $=$= | $\ln a\ f'\left(x\right)\ a^{f\left(x\right)}$lna f′(x) af(x) |
| $\frac{d}{dx}a^{f\left(x\right)}$ddxaf(x) | $=$= | $\ln a\ f'\left(x\right)\ a^{f\left(x\right)}$lna f′(x) af(x) |
Practice question
Question 5
Differentiate $y=5^{2-4x}+3$y=52−4x+3.