Implicit differentiation

  • Thread starter cd246
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Homework Statement


I just got started on this, and am not grasping the WHOLE idea.
1.xy=25 The answer says -y/x
2.x^2+3xy+y^2=15 And this says -y^2/x^2


Homework Equations


1. dy/dx(xy)= dy/dx(25)
1=0 ???
2.dy/dx x^2+3xy+y^2= dy/dx 15
2x+3+y(dy/dx) =0

The Attempt at a Solution


2. -2x-3/y (which i know is not right.)
 

Answers and Replies

  • #2
nrqed
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Homework Statement


I just got started on this, and am not grasping the WHOLE idea.
1.xy=25 The answer says -y/x
2.x^2+3xy+y^2=15 And this says -y^2/x^2


Homework Equations


1. dy/dx(xy)= dy/dx(25)
1=0 ???
2.dy/dx x^2+3xy+y^2= dy/dx 15
2x+3+y(dy/dx) =0

The Attempt at a Solution


2. -2x-3/y (which i know is not right.)

The basic idea is this: Consider y as a function of x.

For example, you have x times y(x) = 25.

Now differentiate both sides with respect to x. You may use the chain rule on the product x times y(x).

[tex] \frac{d}{dx} ( x y(x)) = \frac{d}{dx} (25) [/tex]

Apply the chain rule, take the derivatives and then isolate dy/dx.
 
  • #3
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I don't know if I got this right.
I used the product rule: (x)(y(x))'+(x)'(y(x)) and for derivitive of y(x), I used the
chain rule to find the derivitive. And got(x)(1*x*x)+(x)(y(x)). -> x^3+xy(x)d/dx=0 What else am I doing wrong?
 
  • #4
nrqed
Science Advisor
Homework Helper
Gold Member
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I don't know if I got this right.
I used the product rule: (x)(y(x))'+(x)'(y(x)) and for derivitive of y(x), I used the
chain rule to find the derivitive. And got(x)(1*x*x)+(x)(y(x)). -> x^3+xy(x)d/dx=0 What else am I doing wrong?

You can't use the chain rule to find the derivative of y(x) since it's an unknown function. Just leave it as [itex] y' [/itex]! Now, using (x)' = 1, you get

y + x y' = 0 so y' = -y/x

It's that simple!
 

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