Product rule help

**Paulo1913** · May 17th 2009, 07:08 PM

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....

**TheEmptySet** · May 17th 2009, 07:16 PM

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

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

**Paulo1913** · May 17th 2009, 07:19 PM

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

**VonNemo19** · May 17th 2009, 07:44 PM

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

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

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

**Paulo1913** · May 17th 2009, 07:47 PM

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

**VonNemo19** · May 17th 2009, 08:04 PM

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

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.

**VonNemo19** · May 17th 2009, 08:07 PM

where did that extra 1/2 come from?

Here

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

now thats that

**Paulo1913** · May 17th 2009, 08:59 PM

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

(2x-3)(sqrtx)

**VonNemo19** · May 18th 2009, 09:51 AM

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.......

Thread: Product rule help

LinkBack

Thread Tools

Display

Product rule help

Similar Math Help Forum Discussions

[SOLVED] quick question on product rule and equality rule for logs

Finding dy/dx of equation using both chain rule and product rule

Derivatives - product rule + chain rule

Help with a question involving chain rule + product rule (derivative)

Intergration help, reverse chain rule, product rule

Search Tags