y=2x +200 - √(x^2 -300)
y=2x +200 -(x^2-300)^½
y'=2 - (2x)(½)(x^2 -300)^-½
y'=2 - x(x^2 -300)^-½
y'=2 - x/√(x^2 - 300)
2=x/√(x^2 - 300)
x=2(√(x^2 - 300)

I'm lost from here =[

Quote:

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Originally Posted by Eelarch

y=2x +200 - √(x^2 -300)
y=2x +200 -(x^2-300)^½
y'=2 - (2x)(½)(x^2 -300)^-½
y'=2 - x(x^2 -300)^-½
y'=2 - x/√(x^2 - 300)

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So far so good... in fact, you're done! The answer is indeed:



** Start EDIT ** I'm afraid I made a mistake here - there should not be a 2 in the denominator. It should be:


My Apologies for that error ! **End EdIt**

Quote:

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Originally Posted by Eelarch

2=x/√(x^2 - 300)

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Huh? What happened to the y'? If you want to continue rearranging things, you would have:



But what's the point? You've aleady got your answer.

**EDIT - again., the 2 in the denominator should not be there.

Are you sure? I vaugely remember my maths teacher getting a nice answer by factorising the bracket in some way, this might seem rude so I apologise but could you check over the working just to make sure no nice whole answer can come out of it?

1/4 = (x^3 -300x)/1

1=4(X^3 -300x)
1= 4x(x^2 -300)

or am I digging a hole? :P

First - please note that I had an eror in my original response - I had an extra 2 in the denominator, which I have since fixed. Again - your original answer for y' is correct.

I'm not sure where you're heading. You could "clean up" the term that has the square root in the denominator by multiplying both numerator and denominator by that square root:



Or you could show the equation using a negative exponent:



Beyond that, I don't see how further manipulations help anything here..

ah okay, well thank you very much for you're your help , I was pretty confused about it =]