Differentiate 80*[sqrt (500^2+x^2)] + 50*(800-x)

**dwat** · May 17th 2009, 03:42 PM

Differentiate 80*[sqrt (500^2+x^2)] + 50*(800-x).

I've started by

80*[sqrt (500^2+x^2)] + 50*(800-x)

=80* (250000 + x^2)^(1/2) + 50*(800-x)

Now that I've set the question up to differentiate, I don't know what the next step would be if I was to use the product rule

Would it look something like this

Using the Chain Rule for:80* (250000 + x^2)^(1/2)

40(250000+x^2)^(-1/2) * 2x

Again I am not sure of my workings if its correct on this one

Please. Can someone walk me through this problem step by step so I understand how you can derive this function??

Thanks

**VonNemo19** · May 17th 2009, 04:00 PM

here's whatcha got

$\frac{d}{dx}[80(500^2+x^2)^{1/2}+50(800-x)]$
Her I will use the chain rule and the distributive law and make use of the fact that the derivative of a sum is the sum of the derivatives
$=40(500^2+x^2)^{-1/2}2x+\frac{d}{dx}[50(80)-50x]$

Now since 50(80) is constant Its derivative is 0 and I finish up with the power rule

$=40(500^2+x^2)^{-1/2}2x+\frac{d}{dx}[50(80)-50x]=80x(500^2+x^2)^{-1/2}-50$

Does that help?

I think that you were focused on things that you shouldn't have been. Remember, they want you to find the derivative, so don't focus on tedious multiplications, rather look for ways to get rid of those nasty little constants.

**dwat** · May 18th 2009, 01:58 AM

Cool, you are right. I was carried away with the little things such as expanding brackets with powers and fractions. It just got too much and confused me but looking at how you differentiate things, I kinda get a little glare of what differentiation is all about.

That a term dont neccessarily mean you have to go through and multiply a term to the whole equation. Thanks heaps

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