# What is the derivative of sinx/(x^3-2x)?

Jul 28, 2015

${y}^{'} = \frac{1}{{x}^{3} - 2 x} \cdot \cos x - \frac{3 {x}^{2} - 2}{{x}^{3} - 2 x} ^ 2 \cdot \sin x$

#### Explanation:

The first important thing to notice here is that your function is actually the quotient of two other functions, which means that you can use the quotient rule to differentiate it.

More specifically, these two functions are

$\left\{\begin{matrix}f \left(x\right) = \sin x \\ g \left(x\right) = {x}^{3} - 2 x\end{matrix}\right.$

Your function can thus be written as

$y = f \frac{x}{g} \left(x\right)$

The quotient rule tells you that you can differentiate a function that is the quitient of two other functions by

d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/([g(x)]^2, where $g \left(x\right) \ne 0$

Another important thing to remember is that

$\frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos x$

So, the derivative of your function $y$ will look like this

${y}^{'} = \frac{\left(\frac{d}{\mathrm{dx}} \left(\sin x\right)\right) \cdot \left({x}^{3} - 2 x\right) - \sin x \cdot \frac{d}{\mathrm{dx}} \left({x}^{3} - 2 x\right)}{{x}^{3} - 2 x} ^ 2$

${y}^{'} = \frac{\cos x \cdot \left({x}^{3} - 2 x\right) - \sin x \cdot \left(3 {x}^{2} - 2\right)}{{x}^{3} - 2 x} ^ 2$

This is equivalent to

${y}^{'} = \frac{\cos x \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\left({x}^{3} - 2 x\right)}}}}{{x}^{3} - 2 x} ^ \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} - \frac{\sin x \cdot \left(3 {x}^{2} - 2\right)}{{x}^{3} - 2 x} ^ 2$

${y}^{'} = \textcolor{g r e e n}{\frac{1}{{x}^{3} - 2 x} \cdot \cos x - \frac{3 {x}^{2} - 2}{{x}^{3} - 2 x} ^ 2 \cdot \sin x}$