Question

How do you find the derivative of `f(x)=2/root3(x)+3cosx`?

Answer 1

`-2/(3root(3)x^4)-3sinx`

Explanation:

First, we can write those square root/ other roots into a simple form `x^n`.
Then we can apply the power rule:
`y=x^n`
`dy/dx=y'=nx^(n-1)`

Let's start to solve it!
`f(x)=2/root(3)x+3cosx`
`=2x^(-1/3)+3cosx`

`f'(x)=2(-1/3)x^(-1/3-1)+3(-sinx)`
`=-2/3x^(-4/3)-3sinx`
`=-2/(3root(3)x^4)-3sinx`

Here is the answer. Hope it can help you :)
Feel free to ask me if you have any questions.

Answer 2

Please see below.

Explanation:

Use the power rule and the derivative of cosine.

`d/dx(2/root(3)x) = d/dx(2x^(-1/3)) = -2/3x^(-4/3) = -2/(3x^(4/3))`

`d/dx(3cosx) = 3(-sinx) = -3sinx`.

So,

`f'(x) = -2/3x^(-4/3)-3sinx`

Rewrite the first term to preference.