Derivative of y = sqrt(x)(x - 1): where does the 3 come from?

[communitycoll]

I'll just make one thread for all the help I'll need with derivatives so I don't clutter up this forum.

Homework Statement

Find the derivative of y = sqrt(x)(x - 1).

Homework Equations

Wolfram Alpha gets this:
http://www.wolframalpha.com/input/?i=derivative+y+=+sqrt(x)(x+-+1)

I got sqrt(x) + [(x - 1) / (2sqrt(x))]. Which is basically everything up until the point where Wolfram returns the answer.

I don't understand where the 3 in the numerator comes from, or where the sqrt(x) that's being added goes.

The Attempt at a Solution

All the work you see Wolfram doing, up until the point Wolfram returns the answer.

 

[HallsofIvy]

It appears that you have used the product rule: the derivative of $x^{1/2}(x-1)$ is $(1/2)x^{-1/2}(x- 1)+ x^{1/2}(1)= \sqrt{x}+ (x- 1)/2\sqrt{x}$.

However, you can also write $x^{1/2}(x- 1)= x^{3/2}- x^{1/2}$. Then the derivative is $(3/2)x^{1/2}- (1/2)x^{-1/2}$. That is what Wolfram is doing.

Of course, those are the same. In the your answer, $\sqrt{x}+ (x-1)/2\sqrt{x}$, $x/\sqrt{x}= \sqrt{x}$ so that can be written $\sqrt{x}+ (1/2)\sqrt{x}- 1/2\sqrt{x}= (3/2)\sqrt{x}- (1/2)x^{-1/2}$, the same as Wolfram's answer.

 

[eumyang]

Wolfram added the fractions together by finding the LCD.
$$\sqrt{x} + \frac{x - 1}{2\sqrt{x}}$$
The 1st "fraction" has a denominator of 1, so the LCD is 2 sqrt (x). Multiply top and bottom of the 1st "fraction" by this LCD:
$$\frac{\sqrt{x} \cdot 2\sqrt{x}}{1\cdot 2\sqrt{x}} + \frac{x - 1}{2\sqrt{x}}$$
I'll let you figure out the rest.