6.07 Differentiation and logarithms

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!

 

As the natural logarithm function $y=\log_ex$y=loge​x 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=loge​x 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.

 

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

 

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)​

 

 

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

Question 3

 

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.

 

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

Question 5