New Zealand
Level 8 - NCEA Level 3

# Basic derivative of Logs

Lesson

## The derivative of a natural log function

Recall from our work on the calculus of exponential functions that if $y=e^x$y=ex then $\frac{dy}{dx}=e^x$dydx=ex.

From the definition of $y=\log_ex$y=logex we know that $x=e^y$x=ey, and so, following the pattern above, we can conclude that $\frac{dx}{dy}=e^y$dxdy=ey

Inverting the elements in this equation we can then say that $\frac{dy}{dx}=\frac{1}{e^y}$dydx=1ey, but since $x=e^y$x=ey, we conclude that $\frac{dy}{dx}=\frac{1}{x}$dydx=1x.

We can write our result using the natural logarithm notation as:

$\frac{d}{dx}\left(\ln x\right)=\frac{1}{x}$ddx(lnx)=1x

This is a fascinating result. The gradient of the tangent at the point $x=a$x=a to the curve given by $y=\ln x$y=lnx  is simply $\frac{1}{a}$1a

Have a play with this applet and see for yourself.

At $x=1$x=1 for example, $\frac{dy}{dx}=\frac{1}{1}=1$dydx=11=1. At $x=2$x=2$\frac{dy}{dx}=\frac{1}{2}$dydx=12. At $x=3$x=3$\frac{dy}{dx}=\frac{1}{3}$dydx=13 and so on. As $x$x becomes larger and larger the tangents gradient diminishes, but always remaining positive as it does so. The gradient at $x=1000$x=1000 for example is $\frac{dy}{dx}=0.001$dydx=0.001

### Outcomes

#### M8-11

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

#### 91578

Apply differentiation methods in solving problems