Partial derivative

  • Thread starter galipop
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  • #1
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Hi All,

Can someone refresh my memory and show me how to find the following partial derivate:

[tex]t=\frac{x}{\sqrt{x^2+y^2}}[/tex]

with respect to x.

Thanks
 
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Answers and Replies

  • #2
arildno
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The partial derivative of t(x,y) with respect to x is found by differentiating the expression as in 1-variable calculus; just remember that y is to be considered a constant. For example:
[tex]t(x,y)=xy\rightarrow\frac{\partial{t}}{\partial{x}}=y[/tex]
 
  • #3
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Yeah I understand the concept. It's just that I've forgotten how to find the derivative of a quotient. The textbook that I have doesn't explain it clearly.
 
  • #4
arildno
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The simplest way is to solve for the derivative of t w.r. to x by differentiating the equivalent equation with respect to x:
[tex]\sqrt{x^{2}+y^{2}}t(x,y)=x[/tex]

Another easy way is (I'll give it for one variable):
[tex]t(x)=\frac{f(x)}{g(x)}[/tex]
Use the product rule and the chain rule:
[tex]\frac{dt}{dx}=\frac{f'(x)}{g(x)}-\frac{f(x)}{g^{2}(x)}*g'(x)[/tex]
 
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  • #5
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I remember the quotient rule as "Lo d Hi minus Hi d Lo over Lo squared," or:

[tex]\frac{d}{dx}\,\frac{f(x)}{g(x)}=\frac{gf' - fg'}{g^2}[/tex]
 
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  • #6
NateTG
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Personally I generally use:
[tex]\frac{a}{b}=a (b)^{-1}[/tex]
so in your case
[tex]\frac{x}{\sqrt{x^2+y^2}}=x (x^2+y^2)^{-\frac{1}{2}}[/tex]
 
  • #7
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galipop said:
Yeah I understand the concept. It's just that I've forgotten how to find the derivative of a quotient. The textbook that I have doesn't explain it clearly.
When I forget how do calculate a particular derivate I always derive it by using the definition of the derivate, i.e.
[tex]\frac{d}{dx}f(x) = \lim_{h \to \infty} \frac{f(x+h)-f(x)}{h}[/tex]​
e(ho0n3
 
  • #8
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NateTG said:
Personally I generally use:
[tex]\frac{a}{b}=a (b)^{-1}[/tex]
so in your case
[tex]\frac{x}{\sqrt{x^2+y^2}}=x (x^2+y^2)^{-\frac{1}{2}}[/tex]
How do you proceed using this form?
 
  • #9
Product rule and power rule.

f'*g + g'*f, where f = x and g = (x^2 + y^2)^(-1/2), so f' = 1 and g' = -x(x^2 + y^2)^(-3/2)

And e(ho0n3, don't you mean limit as h goes to 0?

cookiemonster
 
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  • #10
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cookiemonster said:
Product rule and power rule.

f'*g + g'*f, where f = x and g = (x^2 + y^2)^(-1/2), so f' = 1 and g' = -x(x^2 + y^2)^(-3/2)

And e(ho0n3, don't you mean limit as h goes to 0?

cookiemonster
Thanks Cookiemonster.
 

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