Question

How do you solve the derivative of (2x^2+4x-3)/sqrt(2x-1) ? Could you please write out the steps to get to the result?

Answer 1

(6 x^2-1)/(2 x-1)^(3/2)

Answer 2

Possible derivation:
d/dx((2 x^2+4 x-3)/sqrt(2 x-1))
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = 1/sqrt(2 x-1) and v = 2 x^2+4 x-3:
(d/dx(2 x^2+4 x-3))/sqrt(2 x-1)+(2 x^2+4 x-3) (d/dx(1/sqrt(2 x-1)))
Using the chain rule, d/dx(1/sqrt(2 x-1)) = -(( du)/( dx))/(2 u^(3/2)), where u = 2 x-1 and ( d1/sqrt(u))/( du) = -1/(2 u^(3/2)):
(2 x^2+4 x-3) (-(d/dx(2 x-1))/(2 (2 x-1)^(3/2)))+(d/dx(2 x^2+4 x-3))/sqrt(2 x-1)
Differentiate the sum term by term and factor out constants:
(2 (d/dx(x^2))+d/dx(-3)+4 (d/dx(x)))/sqrt(2 x-1)-((2 x^2+4 x-3) (d/dx(2 x-1)))/(2 (2 x-1)^(3/2))
Differentiate the sum term by term and factor out constants:
(2 (d/dx(x^2))+4 (d/dx(x))+d/dx(-3))/sqrt(2 x-1)-((2 x^2+4 x-3) (d/dx(-1)+2 (d/dx(x))))/(2 (2 x-1)^(3/2))
The derivative of -3 is zero:
(2 (d/dx(x^2))+4 (d/dx(x))+0)/sqrt(2 x-1)-((2 x^2+4 x-3) (2 (d/dx(x))+d/dx(-1)))/(2 (2 x-1)^(3/2))
The derivative of -1 is zero:
(2 (d/dx(x^2))+4 (d/dx(x)))/sqrt(2 x-1)-((2 x^2+4 x-3) (2 (d/dx(x))+0))/(2 (2 x-1)^(3/2))
The derivative of x is 1:
(2 (d/dx(x^2))+1 4)/sqrt(2 x-1)-((2 x^2+4 x-3) (d/dx(x)))/(2 x-1)^(3/2)
The derivative of x is 1:
(2 (d/dx(x^2))+4)/sqrt(2 x-1)-(2 x^2+4 x-3)/(2 x-1)^(3/2)
The derivative of x^2 is 2 x:
Answer: |
| (2 (2 x)+4)/sqrt(2 x-1)-(2 x^2+4 x-3)/(2 x-1)^(3/2)