We have seen that the exponential function $y=e^x$y=ex has the remarkable property that the gradient of the tangent at any point along the function is equal to the value of the function at that point. That is, the function is its own derivative!
| $\frac{d}{dx}e^x$ddxex | $=$= | $e^x$ex |
As the natural logarithm function $y=\log_ex$y=logex is the inverse function $y=e^x$y=ex it may come as no surprise that it also has an interesting derivative. The graph of $y=\log_ex$y=logex has the property that the gradient of the tangent at any point has a value equal to the reciprocal of the $x$x-value at that point, as depicted the diagram below.
| $\frac{d}{dx}\ln x$ddxlnx | $=$= | $\frac{1}{x}$1x |
As the function $\ln x$lnx is only defined for $x>0$x>0, the domain of the derivative is also restricted to $x>0$x>0.
Differentiating $y=\ln x$y=lnx
In order to prove the derivative of $y=\ln x$y=lnx stated above, we can 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 |
Then using the following property:
| $\frac{dy}{dx}$dydx | $=$= | $\frac{1}{\frac{dx}{dy}}$1dxdy |
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 |
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.
Practice question
Question 1
Consider the function $f\left(x\right)=\ln x$f(x)=lnx.
Determine $f'$f′$\left(5\right)$(5).
Why is $f'$f′$\left(-3\right)$(−3) undefined?
The function is only defined for $x>0$x>0. Therefore the gradient function is undefined for $x\le0$x≤0.
AThe function is defined for all values of $x$x, but the gradient function is only defined when $\frac{1}{x}$1x is positive.
BWhen $x=-3$x=−3, the function has a vertical tangent, so the derivative is undefined.
CThe derivative of $f\left(x\right)$f(x) is $f'\left(x\right)=\frac{1}{x}$f′(x)=1x. At $x=-3$x=−3, $-\frac{1}{3}$−13 is undefined.
D
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{1}{u}\times u'$1u×u′ | |
| $=$= | $\frac{f'\left(x\right)}{f\left(x\right)}$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 examples
Example 1
Find the derivative of $y=2\ln\left(1-x\right)$y=2ln(1−x).
Think: First note that the argument of the log function implies that $x<1$x<1. This is a function in the form $y=\ln\left(f\left(x\right)\right)$y=ln(f(x)). Hence, to find $\frac{dy}{dx}$dydx, we take the derivative of $f\left(x\right)$f(x) and divide it by $f\left(x\right)$f(x).
Do:
| $y$y | $=$= | $y=2\ln\left(1-x\right)$y=2ln(1−x) |
| $\frac{dy}{dx}$dydx | $=$= | $2\left(\frac{-1}{1-x}\right)$2(−11−x) |
| $=$= | $\frac{-2}{1-x}$−21−x | |
| $=$= | $\frac{2}{x-1}$2x−1 |
Reflect: Even though the function $y=\frac{2}{x-1}$y=2x−1 is defined for all $x\ne1$x≠1, the domain of the original logarithmic function restricts the derivative to the interval $x<1$x<1. It is important to consider when interpreting the derivative that the derivative cannot exist outside the natural domain of the function.
Example 2
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 $\frac{dy}{dx}$dydx, 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 questions
Question 2
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.
Question 3
Find the derivative of $\ln\left(\sin x\right)$ln(sinx).
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 3
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) |
| $=$= | $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) |
| $=$= | $2\left(\frac{\left(x-1\right)-\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)$2((x−1)−(x+1)(x+1)(x−1)) | |
| $=$= | $\frac{-4}{x^2-1}$−4x2−1 |
Example 4
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 | $=$= | $\ln\left(\left(x+1\right)^{-3}\right)$ln((x+1)−3) |
| $=$= | $-3\ln\left(x+1\right)$−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 4
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 5
Consider the function $y=2\ln\left(x^2+e\right)$y=2ln(x2+e).
Find the derivative of $y$y.
Evaluate the derivative at $x=0$x=0.
Hence, determine the equation of the tangent line to the curve $y=2\ln\left(x^2+e\right)$y=2ln(x2+e) at $x=0$x=0.
Question 6
Find the derivative of $y=6x^4\ln x$y=6x4lnx.
Question 7
Consider the function $f\left(x\right)=\frac{\ln x}{x}$f(x)=lnxx for $x>0$x>0.
Show that the graph of $f\left(x\right)$f(x) has a stationary point at $x=e$x=e. In your working you may let $u=\ln x$u=lnx and $v=x$v=x.
Determine $f''\left(e\right)$f′′(e). In your working you may let $u=1-\ln x$u=1−lnx and $v=x^2$v=x2.
Hence determine whether the stationary point is a maximum or minimum.
Maximum
AMinimum
B