# Math Help - Product rule help

1. ## Product rule help

How do I find the derivative of the following equation using the product rule?

(x^2-3x+2)(sqrt x)

I put it in the form of the product rule:

(x^2-3x+2)((1/2)x^(-1/2))+(2x-3)(sqrtx)

But I don't know how to simplify this with the square root bit....

2. Originally Posted by Paulo1913
How do I find the derivative of the following equation using the product rule?

(x^2-3x+2)(sqrt x)

I put it in the form of the product rule:

(x^2-3x+2)((1/2)x^(-1/2))+(2x-3)(sqrtx)

But I don't know how to simplify this with the square root bit....
In this case the product rule isn't needed lets simply first

remember that

$\sqrt{x}=x^{1/2}$ so we get

$y=(x^2-3x+2)x^{1/2}=x^{\frac{5}{2}}-3x^{\frac{3}{2}}+2x^{\frac{1}{2}}$

Now when we take the derivative we only need the power rule

$\frac{dy}{dx}=\frac{5}{2}x^{\frac{3}{2}}-\frac{9}{2}x^{\frac{1}{2}}+x^{-\frac{1}{2}}$

3. So what would that look like when put back into the original form? ie with roots in it?

4. I'm not sure what your asking. If you mean by 'putting back to the other form' by reversing the diferentiaation process, that's called integraton and all you would have to do is the power rule backwards. If you mean getting rid of the fractional expoents, just factor out the smallest one like so

$\frac{5}{2}x^\frac{3}{2}-\frac{9}{2}x^\frac{1}{2}+x^\frac{-1}{2}$

$(\frac{5}{2}x^\frac{4}{2}-\frac{9}{2}x^{1}+1)x^\frac{-1}{2}=(\frac{5x^2}{2}-\frac{9x}{2}+1)\frac{1}{\sqrt{x}}$

And of course there's ways to rationalize the denominator and find the LCD. But I'm sure you can see that. Has this info been helpful?

I'll show you

$(\frac{5x^2}{2}-\frac{9x}{2}+\frac{2}{2})\frac{1}{\sqrt{x}}\frac{\ sqrt{x}}{\sqrt{x}}$

$=(\frac{5x^2-9x+2}{2})\frac{\sqrt{x}}{x}$

5. Sorry I got confused. What I mean is how do I simplify
(x^2-3x+2)((1/2)x^(-1/2))

6. $x^2-3x+2=(x-1)(x-2)$

and of course $\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}$

so after rationalizing the denominator we have

$(x-1)(x-2)\frac{\sqrt{x}}{2x}$

I don't think it gets simpler than that.

7. where did that extra 1/2 come from?

Here

$(x-2)(x-1)\sqrt{x}$

now thats that

8. Thanks. How do I simplify the 2nd part of the product rule, namely

(2x-3)(sqrtx)

9. Easy! Check it out.......

I'm just gonna do the whole problem for you.

Here's the product rule
$\frac{d}{dx}(uv)=(u)\frac{dv}{dx}+(v)\frac{du}{dx}$
Right? Right!

So then what do you have?

$\frac{d}{dx}[(x^2-3x+2)x^{1/2}]=(x^2-3x+2)\frac{d}{dx}x^{1/2}+x^{1/2}\frac{d}{dx}(x^2-3x+2)$

and since $\frac{d}{dx}x^{1/2}=\frac{1}{2}x^{-1/2}$ and $\frac{d}{dx}(x^2-3x+2)=2x-3$ (by using the power rule and the fact that the derivative of any constant is zero), We end up with the following

$\frac{1}{2}x^{-1/2}(x^2-3x+2)+x^{1/2}(2x-3)$

Factoring out the least common factor

$x^{-1/2}[\frac{1}{2}(x^2-3x+2)+x(2x-3)]$

simplifying

$x^{-1/2}(\frac{5}{2}x^2-\frac{9}{2}x+1)$

The product rule si good sometimes, but if I were you, I'd try to do problems like these by first multiplying out the expression, and then using the power rule. The product rule is usually reserved for bigger expressions, or expressions that can't be done otherwise.

Peace out homie.......