Implicit Differentiation: Finding Derivatives of an Implicit Function

In summary, the first derivative is x2+2xy-y2-a2=0, and the second derivative is y^{''}=-\frac{x+y}{x-y}.
  • #1
manenbu
103
0

Homework Statement



Assume that the following equation define the implicit function y=(x). Find the its derivative:
x2 + 2xy - y2 = a2
y'=?
y''=?

Homework Equations



[tex]\frac{dy}{dx} = -\frac{F_x}{F_y}[/tex]

The Attempt at a Solution



so for the first derivative I express that equation as F = x2 + 2xy - y2 - a2 = 0 and using the rule from above I get:
[tex]y^{'} = -\frac{x+y}{x-y}[/tex] which is correct.
For the second derivative the answer should be:
[tex]y^{''} = \frac{2a^{2}}{(x-y)^{3}}[/tex]
But I don't understand how to get there. Where did the 2a2 come from? the 3 hints to me that I need to make a derivative of the fraction, but I can't seem to get anything useful.
 
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  • #2
you can just go through and differentiate everything w.r.t. x

[itex]x^2+2xy-y^2=a^2[/itex]

so, remembering to use the product rule on the second term and the chain rule on the third,

[itex]2x+2y+2x \frac{dy}{dx}-2y \frac{dy}{dx}=0[/itex]

then collect [itex]\frac{dy}{dx}[/itex] terms as follows:

[itex]2(x+y)=2(y-x) \frac{dy}{dx} \Rightarrow \frac{dy}{dx}=y'=\frac{x+y}{y-x}[/itex] which is what you get.

just do the same again to get the second derivative.
 
  • #3
Ok got it right this time.
Just a few questions - why do you add the differential to the y and not the x?
And in the chain rule - where is it?
Am I correct thinking that the chain rule should be (dy/dx)(dx/dx) which cancels to dy/dx?
 
  • #4
manenbu said:
Ok got it right this time.
Just a few questions - why do you add the differential to the y and not the x?
And in the chain rule - where is it?
Am I correct thinking that the chain rule should be (dy/dx)(dx/dx) which cancels to dy/dx?
All latentcorpse did, was differentiate the whole expression with respect to x. Explicitly,

[tex]\frac{d}{dx}\left(x^2+2xy-y^2\right)=\frac{d}{dx}a^2[/tex]

[tex]\frac{d}{dx}x^2 + 2\frac{d}{dx}\left(xy\right) - \frac{d}{dx}y^2 = 0[/tex]

Using the product rule on the second term and the chain rule on the final term,

[tex] 2x + 2y\frac{d}{dx}x + 2x\frac{dy}{dx} - \frac{d}{dy}y^2\cdot\frac{dy}{dx} = 0[/tex]

[tex]2x + 2y + 2x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0[/tex]

Do you follow?
 
  • #5
Ok - everything is understood now. Thanks!
 

1. What is implicit differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of a function that is not explicitly written in terms of one variable. It allows us to find the slope of a curve at any point, even if the equation of the curve is not given in the form y = f(x).

2. When is implicit differentiation used?

Implicit differentiation is commonly used when the equation of a curve cannot be easily solved for y in terms of x. This often occurs when the equation involves both x and y and cannot be rearranged into a simple form. It is also used when the derivative of a function is needed but the function itself is too complex to differentiate using traditional methods.

3. How is implicit differentiation different from explicit differentiation?

Explicit differentiation is used to find the derivative of a function that is written explicitly in terms of one variable. This means that the equation is already solved for y and the derivative can be found using the power rule, product rule, chain rule, etc. Implicit differentiation, on the other hand, is used when the equation is not solved for y and requires the use of the implicit differentiation rule.

4. What is the implicit differentiation rule?

The implicit differentiation rule states that to find the derivative of y with respect to x in an implicit equation, we differentiate both sides of the equation with respect to x and then solve for dy/dx. This involves using the chain rule and the product rule as needed.

5. What are some common applications of implicit differentiation?

Implicit differentiation has many applications in physics, engineering, and economics. It is commonly used to find the rates of change of quantities that are related by implicit equations, such as velocity and acceleration, or supply and demand. It is also used in optimization problems, where we need to find the maximum or minimum value of a function.

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