# How do you derive y = (x-1)/sin(x) using the quotient rule?

Nov 6, 2015

The derivative is $\setminus \frac{\sin \left(x\right) - \left(x - 1\right) \cos \left(x\right)}{{\sin}^{2} \left(x\right)}$

#### Explanation:

The rule states that

$\left(f \frac{x}{g} \left(x\right)\right) ' = \setminus \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{{g}^{2} \left(x\right)}$.

So, as long as we know $f \left(x\right) , g \left(x\right) , f ' \left(x\right) , g ' \left(x\right)$ and ${g}^{2} \left(x\right)$, we're ready to write the derivative. Let's compute this quantities:

$f \left(x\right)$ is of course $x - 1$, and $g \left(x\right)$ is $\sin \left(x\right)$.

This means that $f ' \left(x\right) = 1$, and $g ' \left(x\right) = \cos \left(x\right)$.

Finally, ${g}^{2} \left(x\right) = {\sin}^{2} \left(x\right)$

Now we have all the "ingredients", and we can put them in the "recipe":

$\setminus \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{{g}^{2} \left(x\right)} \to \setminus \frac{1 \cdot \sin \left(x\right) - \left(x - 1\right) \cos \left(x\right)}{{\sin}^{2} \left(x\right)}$