How do you find the derivative for #sqrt(4x)/(x-1)#?

Question

1869 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-14T16:19:53

Quocient rule time!

Let's just rewrite the two functions the compose your fraction: the first one is the same as #sqrt(2*2*x)#, which is the same as #2sqrt(x)#.

The quocient rule determines that:

If #y=(f(x))/(g(x))#, then #(dy)/(dx)# = #(f'(x)*g(x)-f(x)*g'(x))/(f(x)^2)#.

Now, let's just proceed according to the quocient rule.

#(dy)/(dx) = (1/((x-1)^2)*(x-1) - 2sqrt(x)*1)/(4x)#

#(dy)/(dx) = ((x-1)/(x-1)^2-2sqrt(x))/(4x)#

#(dy)/(dx) = (1/(x-1) - 2sqrt(x))/(4x)#

#(dy)/(dx) = (4x)/(x-1)-sqrt(x)/(2x)#

Finally:

#(dy)/(dx) = (4x)/(x-1)-1/(2sqrt(x)#