Question

What is the derivative of `sqrt(2x)`?

Answer 1

Power rule: `(dy)/(dx)[x^n]=n*x^(n-1)`

Power rule + chain rule: `(dy)/(dx)[u^n]=n*u^(n-1)*(du)/(dx)`

Let `u=2x` so `(du)/(dx)=2`

We're left with `y=sqrt(u)` which can be rewritten as `y=u^(1/2)`

Now, `(dy)/(dx)` can be found using the power rule and the chain rule.

Back to our problem: `(dy)/(dx)= 1/2 * u^(-1/2)*(du)/(dx)`

plugging in `(du)/(dx)` we get:

`(dy)/(dx)= 1/2 * u^(-1/2)*(2)`

we know that: `2/2=1`

therefore, `(dy)/(dx)=u^(-1/2)`

Plugging in the value for `u` we find that:
`(dy)/(dx)=2x^(-1/2)`