Differentiation of Quotients and Higher Derivatives

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  • #1
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1) The line 2x+9=3 meets curve xy+y+2=0 at the points P and Q. Calculate the gradient of the curve at P and Q

2)Given that [itex]y=(x^2)/(x-2)[/itex], find
a) [itex](d^2)y/dx^2[/itex] in its simplest form
b)ther range of value for which [itex]dy/dx[/itex] and [itex](d^2)y/dx^2[/itex] are positive.


I can't figure out either of the sums. For the first one I got the answer -1/11 & -44/81 (even though the answer page showed 1/2 and 4/81) and the second I couldn't do. Can anyone do them step by step?
 

Answers and Replies

  • #2
ehild
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What is the first derivative of ##y=\frac{x^2}{x-2}##

ehild
 
  • #3
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y=(x^2)-4x/(x-2)^2
 
  • #4
Ray Vickson
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y=(x^2)-4x/(x-2)^2

How did you get
[tex] \frac{d}{dx} \frac{x^2}{x-2} = x^2 - \frac{4x}{(x-1)^2} ? [/tex]
This is obviously wrong.
 
  • #5
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No, i got [itex]y= \frac{x^2-4x}{(x-2)^2} [/itex]
 
  • #6
HallsofIvy
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1) The line 2x+9=3 meets curve xy+y+2=0 at the points P and Q. Calculate the gradient of the curve at P and Q
Do you mean "2x+ 9y= 3"? What are P and Q?

2)Given that [itex]y=(x^2)/(x-2)[/itex], find
a) [itex](d^2)y/dx^2[/itex] in its simplest form
b)ther range of value for which [itex]dy/dx[/itex] and [itex](d^2)y/dx^2[/itex] are positive.


I can't figure out either of the sums. For the first one I got the answer -1/11 & -44/81 (even though the answer page showed 1/2 and 4/81) and the second I couldn't do. Can anyone do them step by step?
As Ray Vickson pointed out, you have written the first derivative incorrectly- although you may have calculated it correctly. You have the parentheses in the wrong place and you have the denominator wrong.
 
  • #7
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Sorry for that, I suck at typing equations on computer.
 
  • #9
ehild
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No, i got [itex]y= \frac{x^2-4x}{(x-2)^2} [/itex]


Did you mean ##\frac{dy}{dx}##?


ehild
 

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