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

1 Answer
Aug 16, 2015

Answer:

#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))#