# Dy/dx of ln(ax) for any a

1. Nov 30, 2013

It's easy to show that $\frac{dy}{dx}$ of $y = ln(ax)$ where $a \in ℝ, a > 0$ is always $\frac{1}{x}$ :

$y = ln(ax)$
$y = ln(a) + ln(x)$

$ln(a)$ is constant so its derivative is 0, and the derivative of $ln(x)$ is $\frac{1}{x}$.
Hence:

$\frac{dy}{dx} = \frac{1}{x}$

Attached is an image of y=ln(ax) with a = 1,2,3,4,5. It also shows all of their derivatives to be the same (the curve at the top). But they are clearly different curves!!!! There must be some point at which one has a larger gradient than the others?? What am I missing?

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2. Nov 30, 2013

### Staff: Mentor

The curves are just shifted upwards/downwards with respect to each other. This is not easy to see in the graph (as their non-vertical "distance" is not constant), but it is right. And constant offsets do not show up in the derivative.

3. Nov 30, 2013

oh of course!!!! :D I feel a bit stupid now :P I guess they're shifted up by ln(a) right?

4. Nov 30, 2013

### Office_Shredder

Staff Emeritus
That's right.

5. Nov 30, 2013

### Staff: Mentor

Also, you don't "take dy/dx" of some function. dy/dx already represents the derivative of y with respect to x.