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Chapter 3: Problem 18

Find the following derivatives. $$\frac{d}{d x}\left(\ln \left|x^{2}-1\right|\right)$$

Short Answer

Expert verified
Question: Find the derivative of the function $$y = \ln\left|x^{2}-1\right|$$ with respect to x. Answer: The derivative of the function is: $$\frac{d}{d x}\left(\ln\left|x^{2}-1\right|\right)= \begin{cases} \frac{2x}{x^2-1}, & \text{if}\ x^2 - 1 \geq 0 \\ -\frac{2x}{1-x^2}, & \text{if}\ x^2 - 1 < 0 \end{cases} $$

Step by step solution

01

Apply the chain rule

Apply the chain rule for the derivative of the function, which states: $$ \frac{d}{dx} (\ln(u)) = \frac{1}{u} \frac{du}{dx} $$ Here, we have: $$u=\left|x^{2}-1\right|$$ So, we need to find $$\frac{du}{dx}$$.
02

Differentiate the inner function

To differentiate the inner function $$u=\left|x^{2}-1\right|$$, we need to consider two cases as the absolute value function has a piecewise definition: Case 1: When $$x^2 - 1 \geq 0$$ (positive or zero), we have: $$u = x^2 - 1$$ Differentiate this function with respect to x: $$\frac{du}{dx} = 2x$$ Case 2: When $$x^2 - 1 < 0$$ (negative), we have: $$u=-(x^2-1)$$ Differentiate this function with respect to x: $$\frac{du}{dx} = -2x$$ Therefore, the derivative of the inner function is $$\frac{du}{dx} = \pm 2x$$ where the positive value is for $$x^2 - 1 \geq 0$$ and the negative value is for $$x^2 - 1 < 0$$.
03

Apply the chain rule to find the total derivative

Now find the total derivative using the chain rule: $$\frac{d}{dx}\left(\ln\left|x^{2}-1\right|\right)=\frac{1}{\left|x^{2}-1\right|}\cdot (\pm 2x)$$ However, since absolute value function has a piecewise definition, we should write the final result in two cases. So, the derivative of the given function is: $$\frac{d}{d x}\left(\ln\left|x^{2}-1\right|\right)= \begin{cases} \frac{2x}{x^2-1}, & \text{if}\ x^2 - 1 \geq 0 \\ -\frac{2x}{1-x^2}, & \text{if}\ x^2 - 1 < 0 \end{cases} $$

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