This one I just don't know how to derive.

$\ln\|x^4\cos x\|$

I know the derivative of $\ln\ x$, is just $\frac{1}{x}$ . It is the absolute value that throws me off. My question is, does the absolute value stay as is or does it disappear?

  • $\frac{1}{|x^4\cos x|}$
  • $\frac{1}{x^4\cos x}$

Or is there an extra step that I should preform?


4 Answers 4


You can actually show that the derivative of $\ln{|x|}$ is $\frac{1}{x}$ for all $x\ne 0$. For $x>0$ this should be clear; for $x<0$, we know $|x| = -x$, and hence we want to calculate $$\frac{d}{dx}(\ln(-x)) = \frac{1}{-x}(-1) = \frac{1}{x}. $$ Once you know that, then you can proceed with the chain rule, as usual.

  • $\begingroup$ Actually you can derive a more general result. If the derivative of $f(x)$ is odd then the same sequence of calculations that you have shown proves that the derivative of $f(|x|)$ equals the derivative of $f(x)$: $$\frac{d}{dx}(f(-x)) = -\frac{d}{dx}(f(x)) \cdot (-1) = \frac{d}{dx}(f(x))$$. $\endgroup$
    – Bakuriu
    Commented Aug 14, 2015 at 11:36



for $x\ne0$. Now, use the chain rule, but make sure that $x^4 \cos x\ne0$


Let $y=x^4\cos x$. Then, we have $$\frac{d}{dx}\log|x^4\cos x|=\frac{d\log |y|}{d|y|}\text{sgn(y)}\frac{dy}{dx}=\frac{1}{|x^4\cos x|}\text{sgn}(x^4\cos x)(4x^3\cos x-x^4\sin x)=\frac{1}{x^4\cos x}(4x^3\cos x-x^4\sin x)$$Another way to approach the problem is to split it into two parts, (i) $x^4\cos x>0$, and (ii) $x^4\cos x<0$. On Part (i), we have $$\frac{d}{dx}\log|x^4\cos x|=\frac{d}{dx}\log(x^4\cos x)=\frac{1}{x^4\cos x}(4x^3\cos x-x^4\sin x)$$On Part (ii), we have $$\frac{d}{dx}\log|x^4\cos x|=\frac{d}{dx}\log(-x^4\cos x)=\frac{1}{-x^4\cos x}(-4x^3\cos x+x^4\sin x)=\frac{1}{x^4\cos x}(4x^3\cos x-x^4\sin x)$$as expected!

  • $\begingroup$ Sorry, but sgn is unfamiliar to me $\endgroup$
    – sampjr
    Commented Aug 14, 2015 at 4:56
  • $\begingroup$ $\text{sgn}(x)$ is the sign function. It is $1$ for $x>0$ and $-1$ for $x<0$. $\endgroup$
    – Mark Viola
    Commented Aug 14, 2015 at 4:58

To elaborate on Dr. MV's answer, we can find the derivative of the absolute value function by noting $$ |x|=\sqrt{x^2}$$ and then using the chain rule. The proof goes:

$$ \frac d{dx} \sqrt{x^2}=\frac1{2\sqrt{x^2}}\cdot \frac{d}{dx}x^2=\frac{2x}{2\sqrt{x^2}}=\frac{x}{|x|}$$

Now just note that $\frac{x}{|x|}=-1$ if $x<0$ and $\frac{x}{|x|}=1$ for $x>0$.


These kinds of problems can be solved using a simple rule for derivatives of logarithms of absolute functions.

Theorem: For any (univariate) differentiable function $f$ we have:

$$\frac{d}{dx} \ln |f(x)| = \frac{f'(x)}{f(x)} \quad \quad \quad \text{for } f(x) \neq 0.$$

Proof: Since $f$ is differentiable it is continuous, and so, for any point $x$ with $f(x) \neq 0$ there exists a neighbourhood $\mathcal{N}(x)$ where all values of the function have the same sign as $f(x)$ --- i.e., we have: $$\text{sgn} f(x) = \text{sgn} (y) \quad \quad \quad \text{for all } y \in \mathcal{N}(x).$$ Thus, if $f(x) > 0$ then $|f(x)| = f(x)$ over all points in $\mathcal{N}(x)$, and so we have: $$\frac{d}{dx} \ln |f(x)| = \frac{d}{dx} \ln f(x) = \frac{f'(x)}{f(x)}$$ Contrarily, if $f(x) < 0$ then $|f(x)| = -f(x)$ over all points in $\mathcal{N}(x)$, and so we have: $$\ \ \frac{d}{dx} \ln |f(x)| = \frac{d}{dx} \ln (-f(x)) = \frac{-f'(x)}{-f(x)} = \frac{f'(x)}{f(x)}. \ \ \blacksquare$$

Applying this general rule to the case where $f(x) = x^4 \cos x$ we have:

$$\frac{d}{dx} \ln |f(x)| = \frac{4 \cos x - x \sin x}{x \cos x} \quad \quad \quad \text{for all } x \neq \cdots, - \frac{3 \pi}{2}, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{3 \pi}{2}, \cdots.$$


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