Question

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

Answer 1

`y'=-1/x^2-3cos(x)`

Explanation:

`y=1/x-3sin(x)`

You may find it easier to rewrite this as `y=x^-1-3sin(x)`.

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 `(-1)` , then reduce the power by one.

`=>d/dx(x^-1)=(-1)x^-2=-x^-2`

This is equivalent to `-1/x^2`.

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

`=>d/dx(3sin(x))=3cos(x)`

So, we get:

`y'=-1/x^2-3cos(x)`