# find derivative `(sqrt(x))(x-1)` ? Would this just be `1/2x^(-1/2)` or would I need to apply the product rule? Could this be considered the Constant Multiple rule?

### 2 Answers | Add Yours

You *must* to use product rule. If you wish you can first multiply the expression like this

`sqrt x (x-1)=x sqrt x- sqrt x`

and then find the derivative, but this is *not necessary, *it only makes differentiation more difficult.

Let's now differentiate by using product rule.

`(sqrt x(x-1))'=sqrt x'(x-1)+sqrt x(x-1)'=(x-1)/(2sqrt x)+sqrt x`

you have two ways

`A)` first make products so to get:

`sqrt(x)(x-1)= xsqrt(x)-sqrt(x)` and then makes derivatres

`d/dx(xsqrt(x)-sqrt(x))=d/dx xsqrt(x) -d/dx sqrt(x)=` `d/dx x^(3/2) -d/dx x^(1/2)=`

`=3/2x^(3/2-1) -1/2 x^(1/2-1)= 3/2 x^(1/2)-1/2 x^(-1/2)` `=1/2(3sqrt(x)-1/sqrt(x))=`

`1/(2sqrt(x))(3x-1)` `=sqrt(x)/(2x)(3x-1)`

`B)` Soon makes derivative of products:

`d/dx sqrt(x)(x-1)= (x-1) d/dx sqrt(x) + sqrt(x) d/dx (x-1) =`

`=1/2 (x-1)/sqrt(x) + sqrt(x)`

the two results look like differents, instead, still working on the last one :

`1/2 (x-1)/sqrt(x) +sqrt(x)= 1/(2sqrt(x))(x-1+2x)=sqrt(x)/(2x)(3x-1)`

The result is the same.