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Second Derivative of an Implicit function

  • Thread starter Ki-nana18
  • Start date
How would you find the second derivative of an implicit function?
y^2-x^2=16

Heres my attempt:
2y(dy/dx)-2x=0
2y(dy/dx)=2x
2y(dy/dx)/2y=2x/2y
dy/dx= x/y
This is only the first derivative. I think I'm suppose to plug in dy/dx back into the original equation. Am I on the right track?
 
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When my professor taught me implicity, we never really covered higher order derivatives of implicit functions, but here's what I got.

[tex]\frac{dy}{dx}\right)(y^2-x^2=16)=\frac{dy}{dx}\right)=\frac{x}{y}\right)[/tex]

With this implicit function, you can define y explicitly as a function of x and substitute it into the expression.

[tex]y=(16+x^2)^1^/^2[/tex]

Therefore

[tex]\frac{dy}{dx}\right)=\frac{x}{(16+x^2)^1^/^2}\right)[/tex]

Now (i think), you can differentiate again to obtain the second derivative.

[tex]\frac{d^2y}{dx^2}(\frac{x}{(16+x^2)^1^/^2}\right))[/tex]

You don't have to substitute the value of y into the 1st derivative (I think) but it makes differentiating more simple imo.
 
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Hi Ki-nana18, well the first derivative is perfect, however that's not quite the right track from there, lets start from you final equation you wrote:

[tex]
\frac{dy}{dx} = \frac{x}{y}
[/tex]

now what if I rewrote that as:

[tex]
\frac{dy}{dx} = xy^{-1}
[/tex]

and then something that you should know

[tex]
\frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d^{2}y}{dx^2}
[/tex]

meaning that if you take the derivative of a derivative with respect to the same variable (x in this case because its y with respect to x) then it equals the second derivative.

Now ill also say, try thinking about the chain rule and how you could apply that here. Hope that helps Ki-nana :D
 
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When my professor taught me implicity, we never really covered higher order derivatives of implicit functions, but here's what I got.

[tex]\frac{dy}{dx}\right)(y^2-x^2=16)=\frac{dy}{dx}\right)=\frac{x}{y}\right)[/tex]
Granted, you aren't the original poster, but your work deserves some comment.
  1. It makes no sense to take the derivative of an equation. You can take the derivative of each side of an equation, but not the equation itself. What you have written is the derivative dy/dx times an equation being equal to dy/dx, which in turn is equal to x/y.
  2. You are using the symbol dy/dx as if it were the differentiation operator d/dx, which it isn't.
With this implicit function, you can define y explicitly as a function of x and substitute it into the expression.

[tex]y=(16+x^2)^1^/^2[/tex]

Therefore

[tex]\frac{dy}{dx}\right)=\frac{x}{(16+x^2)^1^/^2}\right)[/tex]

Now (i think), you can differentiate again to obtain the second derivative.

[tex]\frac{d^2y}{dx^2}(\frac{x}{(16+x^2)^1^/^2}\right))[/tex]
Here (above) you are writing the 2nd derivative of y with respect to x (which is a function) times an expression instead of the 2nd derivative with respect to x of the same expression. If you confuse d/dx with dy/dx , you will surely come to grief when some equation involves both of these symbols. The same applies to d2/(dx) with d2y/(dx)2.
You don't have to substitute the value of y into the 1st derivative (I think) but it makes differentiating more simple imo.
 

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