Jesse R. answered 06/22/20
BS in Mechanical Engineering with 5+ years tutoring calculus
Hey Eleanor!
Just because the answers you get from one method vs another look different, does not mean that they are not equal.
Here you have a function (x2+2x-1)/x(1/2)
I wrote the square root as an exponent because it helps me when simplifying. If I want to break up the function and take the derivative of each term as I go, having it in this form makes it easier for me.
At this point if you used the quotient rule, you would get ((x(1/2)*(2x +2)) - (1/2)x(-1/2)*(x2+2x-1))/(x(1/2))2 which can then be algebraically simplified.
Instead we will break apart the numerator of the fraction and take the derivative of each term individually.
g(x)= x2/x(1/2) + 2x/x(1/2) - 1/x(1/2)
Using exponent rules, we can simplify this further before taking the derivative
g(x) = x(3/2) + 2x(1/2) - x(-1/2)
Now we take g'(x) and get
g'(x) = (3/2)*x(1/2) + 2*(1/2)*x(-1/2) - (-1/2)*x(-3/2)
g'(x) = (3/2)*sqrt(x) + 1/sqrt(x) +1/(2*sqrt(x3))