I'm having trouble getting to the derivative of $\displaystyle y = x^2 sqrt(x-2)$. Could someone kindly provide a step-by-step solution to the problem? The answer is $\displaystyle x(5x - 8) / 2 sqrt(x-2)$

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- Oct 19th 2009, 01:50 PMArchduke01Derivatives - product rule + chain rule
I'm having trouble getting to the derivative of $\displaystyle y = x^2 sqrt(x-2)$. Could someone kindly provide a step-by-step solution to the problem? The answer is $\displaystyle x(5x - 8) / 2 sqrt(x-2)$

- Oct 19th 2009, 02:17 PMtom@ballooncalculus
Are you sure? Here's a pic - I'll edit details and quite possibly corrections...

http://www.ballooncalculus.org/asy/diffProd/root.png

Actually, the chain rule here is kind of trivial, so we could just use the product rule alone...

http://www.ballooncalculus.org/asy/diffProd/root1.png

... where...

http://www.ballooncalculus.org/asy/prod.png

... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and then, because of the product rule, the whole of the bottom is the derivative of the whole of the top.

Notice I've tweaked the second balloon along the bottom, just for a common denominator.

OK - where does that get us in view of your proposed answer? Have you checked question and answer? Maybe its me...

__________________________________________

Don't integrate - balloontegrate!

http://www.ballooncalculus.org/examples/gallery.html

http://www.ballooncalculus.org/asy/doc.html - Oct 19th 2009, 02:21 PMArchduke01
What's the problem?

If the provided answer caused some confusion, it should be: $\displaystyle x(5x - 8) / (2 sqrt(x-2))$ to be more specific. - Oct 19th 2009, 02:40 PMtom@ballooncalculus
Sorry, should have thought to do an additional tweak for the same reason, then I wouldn't have been confused...

http://www.ballooncalculus.org/asy/diffProd/root2.png

So, yes, there we are. Maybe your mental blip was related to mine. Anyway, hope you like the balloons for tracing through the product rule process - if not, use the formula, of course. - Oct 19th 2009, 03:14 PMArchduke01
- Oct 19th 2009, 03:16 PMtom@ballooncalculus
That was the additional tweak for the sake of the common denominator. (2x because it's the derivative of x^2, then I just doubled the top and bottom.)

Don't be sorry! - find an approach that works for you...

Cheers - Oct 19th 2009, 03:26 PMArchduke01
Ah I finally understand - thank you so much!