# How do you find the derivative of y=1/x-3sinx?

Mar 5, 2017

$y ' = - \frac{1}{x} ^ 2 - 3 \cos \left(x\right)$

#### Explanation:

$y = \frac{1}{x} - 3 \sin \left(x\right)$

You may find it easier to rewrite this as $y = {x}^{-} 1 - 3 \sin \left(x\right)$.

We can take the derivative of each term individually.

First, the derivative of ${x}^{-} 1$. We can use the power rule: multiply the coefficient (in this case $1$) by the power $\left(- 1\right)$ , then reduce the power by one.

$\implies \frac{d}{\mathrm{dx}} \left({x}^{-} 1\right) = \left(- 1\right) {x}^{-} 2 = - {x}^{-} 2$

This is equivalent to $- \frac{1}{x} ^ 2$.

Now we take the derivative of $3 \sin \left(x\right)$. Remember that when we take a derivative, we leave constants alone. The derivative of $\sin \left(x\right)$ is $\cos \left(x\right)$. The inside term $x$ would have a derivative of $1$, so the chain rule isn't necessary.

$\implies \frac{d}{\mathrm{dx}} \left(3 \sin \left(x\right)\right) = 3 \cos \left(x\right)$

So, we get:

$y ' = - \frac{1}{x} ^ 2 - 3 \cos \left(x\right)$