It's easy to show that [itex]\frac{dy}{dx}[/itex] of [itex]y = ln(ax)[/itex] where [itex]a \in ℝ, a > 0[/itex] is always [itex]\frac{1}{x}[/itex] :(adsbygoogle = window.adsbygoogle || []).push({});

[itex]y = ln(ax)[/itex]

[itex]y = ln(a) + ln(x)[/itex]

[itex]ln(a)[/itex] is constant so its derivative is 0, and the derivative of [itex]ln(x)[/itex] is [itex]\frac{1}{x}[/itex].

Hence:

[itex]\frac{dy}{dx} = \frac{1}{x}[/itex]

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

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