1. ## Chain rule help

y = sqrt(2x+1)(x^3)

I always screw this up, so if someone could just walk through the steps so I could compare, I'd really appreciate it.

Thanks.

2. Is it $\displaystyle y = x^{3} \sqrt{2x+1}$?

Then $\displaystyle \frac{dy}{dx} = x^{3} \frac{1}{2}(2x+1)^{-\frac{1}{2}} + 3x^{2} \sqrt{2x+1}$

3. Yeah, that's what I had.

Is it okay to leave it like that or do I have to simplify it more?

Thanks.

4. you can just leave it like that.

5. Originally Posted by tukeywilliams
Is it $\displaystyle y = x^{3} \sqrt{2x+1}$?

Then $\displaystyle \frac{dy}{dx} = x^{3} \frac{1}{2}(2x+1)^{-\frac{1}{2}} + 3x^{2} \sqrt{2x+1}$

Is it okay to leave it like that or do I have to simplify it more?

Thanks.
You can leave it, but technically you have a radical in the denominator:
$\displaystyle \frac{dy}{dx} = x^{3} \frac{1}{2\sqrt{2x + 1}} + 3x^{2} \sqrt{2x+1}$

So
$\displaystyle \frac{dy}{dx} = \frac{x^3 \sqrt{2x + 1}}{4x + 2} + 3x^{2} \sqrt{2x+1}$

It really depends on how you are going to use the derivative from here on.

-Dan

6. Or you can write it as:

$\displaystyle \frac{x^{3}}{2\sqrt{2x+1}} + 3x^{2} \sqrt{2x+1}$