Question

How solve it? Topic: DERIVATE

Please I need help I do not know how to solve

Answer 1

`f(x)=sqrt(2x)`

First, let's rewrite the square root as a `1//2` power.

`f(x)=(2x)^(1/2)`

Now, we need to recognize that these can be split up as a constant and a variable function.

`f(x)=2^(1/2)*x^(1/2)`

When we differentiate, multiplicative constants like the `2^(1/2)` here simply stay on the "outside," that is, we don't do anything to them.

To differentiate `x^(1/2)`, we use the power rule, which says that `d/dxx^n=nx^(n-1)`.

Then, we see that:

`f'(x)=2^(1/2)*(1/2x^(1/2-1))`

Now simplifying:

`f'(x)=2^(1/2)/2x^(-1/2)`

`f'(x)=1/(2^(1/2)*x^(1/2))`

`f'(x)=1/sqrt(2x)`

So at `x=5/3`, the derivative is equal to:

`f'(5/3)=1/sqrt(2*5/3)=1/sqrt(10/3)=sqrt(3/10)`

Answer 2

`f'(x)=2/[2sqrt(2x)]=1/sqrt(2x)`
`f'(5/3)=1/sqrt(10/3)=sqrt3/sqrt10`

Explanation:

show below:

`f(x)=sqrt(2x)`

`f'(x)=2/[2sqrt(2x)]=1/sqrt(2x)`

The derivative at `x=5/3` equal

`f'(5/3)=1/sqrt(10/3)=sqrt3/sqrt10`

`"Note that"`

`color(red)[y=sqrtx]`

`color(red)[y'=1/[2sqrtx]*x']`

`color(red)[sqrt[a/b]=sqrta/sqrtb]`