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Calculus IDerivatives

Logarithms, one simpler step at a time.

Use log laws to untangle products, quotients and changing powers. Learn the method, explore e as a limit, then solve the questions with each step explained.

Derivatives of logarithms · Step 1 of 7

The slope of a natural logarithm is a reciprocal

Burgundy: natural log of absolute x. Blue: tangent at the selected point.

−4−3−2−101234−3−2−10123
The natural logarithm, ln, undoes the exponential with base e: eln⁡x=x{e^{\ln x}} = {x} for positive x. Its derivative is 1x{\frac1x}. As x grows, the graph keeps rising, but its slope gets smaller. A derivative measures that slope at one point.
Try it: Move toward zero on each side. Predict the sign and size of the slope, then reveal it.
Side of zero
Distance from zero1
a=1,f′(a)=11=1.000{\displaystyle a} = {\displaystyle 1,\quad f'(a)} = {\displaystyle \frac1{1}} = {\displaystyle 1.000}
ddxln⁡x=1x(x>0){\displaystyle \frac{d}{dx}\ln x} = {\displaystyle \frac1x\quad(x>0)}
ddxln⁡∣x∣=1x(x≠0){\displaystyle \frac{d}{dx}\ln|x|} = {\displaystyle \frac1x\quad(x\ne0)}

For negative x, absolute x is −x. The chain rule gives the same reciprocal:

ddxln⁡(−x)=−1−x=1x{\displaystyle \frac{d}{dx}\ln(-x)} = {\displaystyle \frac{-1}{-x}} = {\displaystyle \frac1x}

Neither logarithm is defined at zero. Ordinary ln x has no real values on the negative side; ln absolute x does.

Check yourself
On the negative side, is the slope of ln absolute x positive or negative?

Need help untangling a logarithmic derivative?

Bring your practice questions. We work through them together until every type feels routine.

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