The Absolute Value Matters

Calculus Level 2

What is the value of the derivative of $y = \left|\ln(x^2)\right|$ at $x = -\dfrac{1}{2}$ ?

-4 Does Not Exist 2 -2 4

1 solution

Andrew Ellinor
Oct 19, 2015

The derivative of $y = \ln(x)$ is $y' = \dfrac{1}{x}$ .
The derivative of $y = \ln(x^2)$ is $y' = \dfrac{2x}{x^2}$ .

When $x = -\dfrac{1}{2}$ is pluged in, we get -4. However. The absolute value on the outside of $\ln(x^2)$ causes the graph to be reflected above the $x$ -axis making the slope of the tangent line at that point become positive 4.

We could have used the fact that $\frac{d u}{dx} = \frac{u.u'}{|u|}$ to obtain: $\frac{d |\ln (x^2) |}{dx} = \frac{\ln (x^2) \frac{2}{x}}{|\ln (x^2) |}$ and solve for $\ x = - \frac{1}{2}$ .

Leonardo de Araujo - 5 years, 7 months ago

The Absolute Value Matters

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