I've been trying to differentiate $y=\arccos(4/x)$ and ended up with the answer $\frac{dy}{dx} = \frac{4}{x\sqrt{x^2-16}}$ however the solution to this question is marked as $\frac{dy}{dx}= \frac{4}{|x|\sqrt{x^2-16}}$. I opted to use implicit differentiation and honestly have no idea where the $|x|$ could of come from. Any help would be greatly appreciated.
$\endgroup$ 13 Answers
$\begingroup$Clearly,\begin{align}\frac{\mathrm dy}{\mathrm dx}&=\left(-\frac4{x^2}\right)\left(-\frac1{\sqrt{1-\frac{16}{x^2}}}\right)\\&=\frac4{x^2}\cdot\frac1{\sqrt{1-\frac{16}{x^2}}}\end{align}and\begin{align}\frac4{x^2}\cdot\frac1{\sqrt{1-\frac{16}{x^2}}}&=\frac4{|x|}\frac1{|x|\sqrt{1-\frac{16}{x^2}}}\\&=\frac4{|x|\sqrt{x^2-16}}.\end{align}Note that the equality $a\sqrt b=\sqrt{a^2b}$ only holds if $a\geqslant0$.
$\endgroup$ $\begingroup$Using implicit differentiation,
$$\begin{align*} y &= \arccos \frac4x\\ \cos y &= \frac 4x \end{align*}$$
Differentiating both sides with respect to $x$,
$$\begin{align*} -\sin y\cdot \frac{dy}{dx} &= -\frac 4{x^2}\\ \sqrt{1-\cos^2y}\cdot \frac{dy}{dx} &= \frac 4{x^2}\\ \sqrt{1-\frac{16}{x^2}}\cdot \frac{dy}{dx} &= \frac{4}{x^2}\\ \frac{\sqrt{x^2-16}}{\sqrt{{x^2}}}\cdot\frac{dy}{dx} &= \frac{4}{x^2} \end{align*}$$
For $x$ that may be positive or negative, $\sqrt{x^2}$ is really $|x|$, and is different from $x$ when $x$ is negative.
$$\begin{align*} \frac{\sqrt{x^2-16}}{|x|}\cdot\frac{dy}{dx} &= \frac{4}{x^2}\\ \frac{dy}{dx} &= \frac{4}{|x|\sqrt{x^2-16}} \end{align*}$$
$\endgroup$ $\begingroup$The given function $y=\arccos\dfrac{4}{x}$ is not defined as real in $|x|<4$
$$ \text{for } x> 4 , y'=\frac{4}{x\sqrt{x^2-16}},$$$$ \text{for } x< 4 , y'= \pi - \frac{4}{|x|\sqrt{x^2-16}}.$$
