# Thread: Stuff on product rule problem

1. ## Stuff on product rule problem

Question is:

y = (x + 1) \sqrt[x - 1]

So I convert to

y = (x + 1) (x - 1)^(1/2)

Then

dy/dx = (x -1)^(1/2) . 1 + (x + 1) 1/2(x - 1)^(-1/2) . 1

y= (x - 1)^{1/2} + \frac{(x + 1}{2(x - 1)^{\frac{1}{2}}}

Is this correct?

Book says numerator is 3x - 1!

What have I done wrong?

Angus

2. ## Re: Stuff on product rule problem

You haven't done anything wrong, you just need to combine the terms of your answer.

3. ## Re: Stuff on product rule problem

You have given:
$y=(x+1)\cdot \sqrt{x-1}$
which is equal to:
$y=(x+1)\cdot (x-1)^{\frac{1}{2}}$
Using the product rule gives us:
$y'=\frac{d}{dx}(x+1)\cdot (x-1)^{\frac{1}{2}}+(x+1)\cdot\frac{d}{dx}\left[(x-1)^{\frac{1}{2}}\right]$
$y'=(x-1)^{\frac{1}{2}}+\frac{(x+1)\cdot (x-1)^{\frac{-1}{2}}}{2}$
$y'=\sqrt{x-1}+\frac{x+1}{2\sqrt{x-1}}$

Now make a common denominator and add the fractions.

4. ## Re: Stuff on product rule problem

Rationalize the denominator.

5. ## Re: Stuff on product rule problem

Originally Posted by SammyS
Rationalize the denominator.
I did struggle with the rationalising. I was thinking

\sqrt{x-1} * \sqrt{x-1}

would be multiplied like (x - 1)(x - 1) - ie x * x, x * -1 etc, but of course it is not. you just add the powers. ie (x - 1)^(1/2) (x - 1)^(1/2) = (x - 1)

Sorry I will learn how to do latex some time. I put around [tex] tags but it doesn't work for me.

6. ## Re: Stuff on product rule problem

Indeed: $\sqrt{x-1}\cdot \sqrt{x-1}=x-1$ therefore you get:
$y'=\sqrt{x-1}+\frac{x+1}{2\sqrt{x-1}}=\frac{2(x-1)}{2\sqrt{x-1}}+\frac{x+1}{2\sqrt{x-1}}=\frac{2(x-1)+x+1}{2\sqrt{x-1}}=\frac{3x-1}{2\sqrt{x-1}}$

Note:
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