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

1 Answer
Yonas Yohannes
Feb 20, 2016

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

Explanation:

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

Related questions
See all questions in Chain Rule
Impact of this question
1989 views around the world
You can reuse this answer
Creative Commons License