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

1 Answer
Feb 20, 2016

Answer:

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))#