11.16 Differentiating logarithmic functions
Recall that we differentiate the exponential function as follows:
| $\frac{d}{dx}e^x$ddxex | $=$= | $e^x$ex |
We will now use this result to determine how to differentiate the inverse function $y=\ln x$y=lnx.
Differentiating $y=\ln x$y=lnx
In order to find the derivative of $y=\ln x$y=lnx, we want to make use of our knowledge of differentiating exponential functions. To do so, let's first rewrite the equation in exponential form:
| $y$y | $=$= | $\ln x$lnx |
| $e^y$ey | $=$= | $e^{\ln x}$elnx |
| $x$x | $=$= | $e^y$ey |
We can now use our rule for differentiating exponential functions by differentiating $x=e^y$x=ey with respect to $y$y:
| $\frac{dx}{dy}$dxdy | $=$= | $e^y$ey |
As $y=e^x$y=ex and $x=e^y$x=ey are inverse functions we can say that:
| $\frac{dy}{dx}$dydx | $=$= | $\frac{1}{\frac{dx}{dy}}$1dxdy |
And so we have that:
| $\frac{dy}{dx}$dydx | $=$= | $\frac{1}{e^y}$1ey |
Now we can rewrite the derivative back in terms of $x$x by remembering that $x=e^y$x=ey. Thus we have:
| $\frac{dy}{dx}$dydx | $=$= | $\frac{1}{x}$1x |
Therefore for $y=\ln x$y=lnx:
| $\frac{dy}{dx}$dydx | $=$= | $\frac{1}{x}$1x |
| $\frac{d}{dx}\ln x$ddxlnx | $=$= | $\frac{1}{x}$1x for $x>0$x>0 |
Note that $y=\ln x$y=lnx is only defined for $x>0$x>0. As such, the derivative is also only defined over this domain.
Differentiating $y=\ln\left(f\left(x\right)\right)$y=ln(f(x))
Consider functions of the form:
| $y$y | $=$= | $\ln\left(f\left(x\right)\right)$ln(f(x)) |
Letting $u=f\left(x\right)$u=f(x), we have:
| $y$y | $=$= | $\ln u$lnu |
| $\frac{dy}{du}$dydu | $=$= | $\frac{1}{u}$1u |
| $\frac{du}{dx}$dudx | $=$= | $f'\left(x\right)$f′(x) |
Using the chain rule:
| $\frac{dy}{dx}$dydx | $=$= | $\frac{dy}{du}\times\frac{du}{dx}$dydu×dudx |
| $\frac{dy}{dx}$dydx | $=$= | $\frac{1}{u}\times u'$1u×u′ |
| $\frac{dy}{dx}$dydx | $=$= | $\frac{f'\left(x\right)}{f\left(x\right)}$f′(x)f(x) |
| $\frac{d}{dx}\ln\left(f\left(x\right)\right)$ddxln(f(x)) | $=$= | $\frac{f'\left(x\right)}{f\left(x\right)}$f′(x)f(x) for $f\left(x\right)>0$f(x)>0 |
Worked example
Example 1
Differentiate the function $y=\ln\left(2x^3-x+1\right)$y=ln(2x3−x+1).
Think: This is a function in the form $y=\ln\left(f\left(x\right)\right)$y=ln(f(x)). Hence, to find $y'$y′, we take the derivative of $f\left(x\right)$f(x) and divide it by $f\left(x\right)$f(x).
Do: In this case, $f\left(x\right)=2x^3-x+1$f(x)=2x3−x+1. So we obtain the derivative of $y$y as follows:
| $y$y | $=$= | $\ln\left(2x^3-x+1\right)$ln(2x3−x+1) |
| $\frac{dy}{dx}$dydx | $=$= | $\frac{6x^2-1}{2x^3-x+1}$6x2−12x3−x+1 |
Practice question
Question 1
Find the derivative of $y=\ln\left(x^4+2\right)$y=ln(x4+2). You may use the substitution $u=x^4+2$u=x4+2.
Differentiating harder logarithmic functions
When faced with finding the derivative of a more complicated logarithmic function, we can often utilise the logarithmic laws and product and quotient rules to help make the process more manageable.
$\log_bx+\log_by=\log_b\left(xy\right)$logbx+logby=logb(xy)
$\log_bx-\log_by=\log_b\left(\frac{x}{y}\right)$logbx−logby=logb(xy)
$\log_b\left(x^n\right)=n\log_bx$logb(xn)=nlogbx
Worked examples
Example 2
Find the first derivative of $y=\ln\left(\left(\frac{x+1}{x-1}\right)^2\right)$y=ln((x+1x−1)2)
Think: Using the power law and subtraction law for logs to rewrite the function:
| $y$y | $=$= | $2\ln\left(\frac{x+1}{x-1}\right)$2ln(x+1x−1) |
| $y$y | $=$= | $2\left(\ln\left(x+1\right)-\ln\left(x-1\right)\right)$2(ln(x+1)−ln(x−1)) |
Do: This gives us an easier function to differentiate:
| $\frac{dy}{dx}$dydx | $=$= | $2\left(\frac{1}{x+1}-\frac{1}{x-1}\right)$2(1x+1−1x−1) |
| $\frac{dy}{dx}$dydx | $=$= | $\frac{-4}{x^2-1}$−4x2−1 |
Example 3
Differentiate $y=\ln\left(\frac{1}{\left(x+1\right)^3}\right)$y=ln(1(x+1)3)
Think: We can use the power log law to make the function easier to differentiate:
| $y$y | $=$= | $\ln\left(x+1\right)^{-3}$ln(x+1)−3 |
| $y$y | $=$= | $-3\ln\left(x+1\right)$−3ln(x+1) |
Do: This gives us an easier function to differentiate:
| $\frac{dy}{dx}$dydx | $=$= | $\frac{-3}{x+1}$−3x+1 |
Practice questions
Question 2
Consider the function $f\left(x\right)=\ln\left(\sqrt{x^2+1}\right)$f(x)=ln(√x2+1).
Find $f'\left(x\right)$f′(x).
Hence find $f'\left(2\right)$f′(2).
Question 3
Find the derivative of $y=\ln e^{8x}$y=lne8x.
Differentiating $y=\log_ax$y=logax
Functions in the form $y=\log_ax$y=logax, where the logarithm has a base other than $e$e, cannot be differentiated directly. We will first need to make use of the change of base rule for logarithms to rewrite such functions using the natural logarithm.
$\log_ab=\frac{\log_cb}{\log_ca}$logab=logcblogca
Let's use this change of base rule to rewrite the function $y=\log_ax$y=logax in terms of the natural logarithm:
| $y$y | $=$= | $\log_ax$logax |
| $y$y | $=$= | $\frac{\log_ex}{\log_ea}$logexlogea |
| $y$y | $=$= | $\frac{\ln x}{\ln a}$lnxlna |
| $y$y | $=$= | $\left(\frac{1}{\ln a}\right)\ln x$(1lna)lnx |
The expression $\frac{1}{\ln a}$1lna is a constant, which means we can now differentiate the function using the fact that the derivative of $\ln x$lnx is $\frac{1}{x}$1x:
| $\frac{dy}{dx}$dydx | $=$= | $\frac{1}{\ln a}\times\frac{1}{x}$1lna×1x |
| $\frac{dy}{dx}$dydx | $=$= | $\frac{1}{x\ln a}$1xlna |
Worked example
Example 4
Find the first derivative of $y=\log_{10}x$y=log10x
Think: Changing the base we get:
| $y$y | $=$= | $\frac{\ln x}{\ln10}$lnxln10 |
Do: Differentiating:
| $\frac{dy}{dx}$dydx | $=$= | $\frac{1}{x\ln10}$1xln10 |
Practice question
Question 4
Differentiate $y=\log_2x$y=log2x.
Differentiating $y=\log_a\left(f\left(x\right)\right)$y=loga(f(x))
We can use the same process shown above to differentiate more general logarithms of bases other than $e$e - that is, functions in the form $y=\log_a\left(f\left(x\right)\right)$y=loga(f(x)).
Once again, we make use of the change of base rule to rewrite our function in terms of the natural logarithm:
| $y$y | $=$= | $\log_a\left(f\left(x\right)\right)$loga(f(x)) |
| $y$y | $=$= | $\frac{\log_e\left(f\left(x\right)\right)}{\log_ea}$loge(f(x))logea |
| $y$y | $=$= | $\frac{\ln\left(f\left(x\right)\right)}{\ln a}$ln(f(x))lna |
| $y$y | $=$= | $\left(\frac{1}{\ln a}\right)\ln\left(f\left(x\right)\right)$(1lna)ln(f(x)) |
The expression $\frac{1}{\ln a}$1lna is a constant, and we already know that the derivative of $\ln\left(f\left(x\right)\right)$ln(f(x)) is $\frac{f'\left(x\right)}{f\left(x\right)}$f′(x)f(x). Putting this together, we have:
| $\frac{dy}{dx}$dydx | $=$= | $\frac{1}{\ln a}\times\frac{f'\left(x\right)}{f\left(x\right)}$1lna×f′(x)f(x) |
| $\frac{dy}{dx}$dydx | $=$= | $\frac{f'\left(x\right)}{f\left(x\right)\ \ln a}$f′(x)f(x) lna |
Practice question
Question 5
Differentiate $f\left(x\right)=\log_5\left(4x+2\right)-6x$f(x)=log5(4x+2)−6x.