How do you take the derivative of the following function:
$$y=\ln(|\sec(5x) + \tan(5x)|)$$ So $dy/dx$ of the following function.
Thanks in advance!
The steps I have taken don't seem to be correct: $$y=\ln(|\sec(5x) + \tan(5x)|)$$ $$y'={\frac{\frac d{dx}(|\sec(5x) + \tan(5x)|) } {{|\sec(5x)+\tan(5x)| }}}$$ $$y'={\frac{|\sec(5x)*\tan(5x)*5 + \sec^2(5x)*5| } {{|\sec(5x)+\tan(5x)| }}}$$ $$y'={\frac{|\sec(5x)*5[\tan(5x) + \sec(5x)] } {{|\sec(5x)+\tan(5x)| }}}$$
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$\begingroup$Note that if $y=\ln|u(x)|$, then using the Chain Rule, $$\frac{dy}{dx}=\frac{1}{u(x)}\cdot\frac{du}{dx}.\quad \text{See my note below for the explanation.}$$ So, the answer to your question is as follows: $$ \begin{align} \frac{dy}{dx}&=\frac{1}{\sec 5x+\tan 5x}\cdot(5\sec 5x\tan 5x+ 5\sec^25x)\\ &=\frac{1}{\sec 5x+\tan 5x}\cdot 5\sec 5x(\tan 5x +\sec 5x)\\ &=5\sec 5x. \end{align}$$
Note: If $u(x)>0$ then $y=\ln|u(x)|=\ln u(x)$ and so, $$\frac{dy}{dx}=\frac{1}{u(x)}\cdot\frac{du}{dx}.$$ While, if $u(x)<0$ then $y=\ln|u(x)|=\ln [-u(x)]$ and so, $$\frac{dy}{dx}=\frac{1}{-u(x)}\cdot\frac{-du}{dx}=\frac{1}{u(x)}\cdot\frac{du}{dx}.$$
$\endgroup$ 1 $\begingroup$Hint:
The derivative of the absolute function is
$$|x|'=\begin{cases}x<0\to-1\\x=0\to\text{undefined}\\x>0\to1.\end{cases}$$
Use the chain rule.
Alternatively (and simpler),
$$(\ln|x|)'=\frac1x$$ for $x\ne0$.
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