Question

How do you differentiate `y = (x+1) (sqrt (2x-1))`?

Answer 1

The simplified condensed solution is:
`sqrt(2x-1)+(x+1)/sqrt(2x-1)=color(red)((3x)/sqrt(2x-1))`

Explanation:

Let `h(x)=f(x)g(x)` where `f(x)=(x+1) " and " g(x)=sqrt(2x-1)`

Now by the product rule we have:

`(dh(x))/dx=h'(x)=color(blue)(f'(x))g(x) + color(green)(g'(x))f(x)`
`color(blue)(f'(x))=color(blue)(1)`
`color(green)(g'(x))=color(green)((1/(2sqrt(2x-1)))((d(2x-1))/dx)(x+1))`
`color(green)(g'(x) =((x+1)/(sqrt(2x-1)))`

`h'(x) =sqrt(2x-1) + (1/(2sqrt(2x-1)))((d(2x-1))/dx)(x+1)`
`h'(x) =sqrt(2x-1) + ((x+1)/(sqrt(2x-1)))` you can further simplify
`h'(x) = color(red)((3x)/sqrt(2x-1))`