Question

How do you use the product Rule to find the derivative of `x+1) (sqrt(2x-1))`?

Answer 1

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

Explanation:

Since your function can be written as the product of two other functions

`y = underbrace((x+1))_(color(red)(f(x))) * underbrace((2x-1)^(1/2))_(color(green)(g(x)))`

you can differentiate it by using the product rule

`color(blue)(d/dx(y) = [d/dx(f(x))] * g(x) + f(x) * d/dx(g(x)))`

This means that you can write

`d/dx(y) = [d/dx(x+1)] * (2x-1)^(1/2) + (x+1) * d/dx(2x-1)^(1/2)`

You can calculate `d/dx(2x-1)^(1/2)` by using the chain rule for `u^(1/2)`, with `u = (2x-1)`.

`d/dx(u^(1/2)) = d/(du)u^(1/2) * d/dx(u)`

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

`d/dx(2x-1)^(1/2) = 1/color(red)(cancel(color(black)(2))) * (2x-1)^(-1/2) * color(red)(cancel(color(black)(2)))`

Take this back to the calculation of your target derivative to get

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

This can be simplified to give

`y^' = (2x-1)^(-1/2) * (2x - color(red)(cancel(color(black)(1))) + x + color(red)(cancel(color(black)(1))))`

`y^' = (3x)/(2x-1)^(1/2)`

This is equivalent to

`y^' = color(green)((3x)/sqrt(2x-1))`