# Implicit differentiation formula

• Jun 14th 2013, 09:51 AM
norton429
Implicit differentiation formula
Several Calculus books explain Implicit Differentiation by assuming that z is implicitly defined as a function of x and y in F(x,y,z)= 0 equation. Then they derive the formula: dz/dx = -Fx/Fz (note that dx/dz here is a partial derivative). I assumed that the inverse must also be true: dx/dz = -Fz/Fx but this doesn't seem to be true. I don't understand what I am missing (Wondering)

Thanks!
• Jun 14th 2013, 10:19 AM
mnov
Re: Implicit differentiation formula
Quote:

Originally Posted by norton429
Several Calculus books explain Implicit Differentiation by assuming that z is implicitly defined as a function of x and y in F(x,y,z)= 0 equation. Then they derive the formula: dz/dx = -Fx/Fz (note that dx/dz here is a partial derivative). I assumed that the inverse must also be true: dx/dz = -Fz/Fx but this doesn't seem to be true. I don't understand what I am missing (Wondering)

Thanks!

Why do you think that it is not true?
• Jun 15th 2013, 03:58 AM
norton429
Re: Implicit differentiation formula

Here is why I think I am missing something:
I was trying to solve the following problem: find the first partial derivatives of U and V with respect to X and Y and the first partial derivatives of X and Y with respect to U and V, given 2U - V + X^2 + X*Y = 0, U + 2V + X*Y - Y^2 = 0.

To find dU/dX I substituted the V in the second equation from the first equation: V=2U + X^2 + XY => I get eventually: 5U + 2X^2 + 3X*Y - Y^2 = 0. Now, I find Fx = 4X + 3Y, and Fu = 5, thus: dU/dX = -(4X + 3Y)/5, which is the correct answer. Now, if the inverse was true, I should have dX/dU = - 5/(4X + 3Y) but the correct answer seems to be: (4Y - X)/2(X^2 - 2XY -Y^2), and this is why I think I am missing something.

Thanks again!
• Jun 15th 2013, 08:57 AM
HallsofIvy
Re: Implicit differentiation formula
For functions of a single variable, it is true that $\frac{1}{\frac{dx}{dt}}= \frac{dt}{dx}$. For functions of more than one variable it is not that easy.
• Jun 15th 2013, 11:12 AM
mnov
Re: Implicit differentiation formula
Quote:

Originally Posted by HallsofIvy
For functions of a single variable, it is true that $\frac{1}{\frac{dx}{dt}}= \frac{dt}{dx}$. For functions of more than one variable it is not that easy.

I think norton429's 'd' in the original post stood for $\partial$. So he is asking if
$\frac{\partial x}{\partial z} = 1/ \frac{\partial z}{\partial x}$
which is true. Are $\frac{d z}{dx}$ and $\frac{d x}{dz}$ even meaningful quantities here?
• Jun 15th 2013, 04:26 PM
HallsofIvy
Re: Implicit differentiation formula
If z is a function of, say, x and y, then $\frac{\partial z}{\partial x}$ makes sense but what is meant by $\frac{\partial x}{\partial z}$.

If $z= x^2+ y^2$ then $\frac{\partial z}{\partial x}= 2x$. But what, precisely, is meant by $\frac{\partial x}{\partial z}$, since x is NOT a function of z or a function of y and z.
• Jun 15th 2013, 08:37 PM
mnov
Re: Implicit differentiation formula
Quote:

Originally Posted by HallsofIvy
If z is a function of, say, x and y, then $\frac{\partial z}{\partial x}$ makes sense but what is meant by $\frac{\partial x}{\partial z}$.

If $z= x^2+ y^2$ then $\frac{\partial z}{\partial x}= 2x$. But what, precisely, is meant by $\frac{\partial x}{\partial z}$, since x is NOT a function of z or a function of y and z.

Given an equation like $z= x^2+ y^2$, we can view $z = z(x,y)$ to be a function of two independent variables $x$ and $y$ OR $x = x(y,z)$ to be a function of two independent variables $y$ and $z$. The former is used to get $\frac{\partial z}{\partial x} =2x$ and the latter to get $\frac{\partial x}{\partial z} = \frac{1}{2x}$.
• Jun 15th 2013, 09:27 PM
norton429
Re: Implicit differentiation formula
Yes, I meant partial derivatives in this case, just didn't bother to try to use the proper symbols. I thought that this should be a pretty straightforward problem, and I am a bit frustrated to be stuck on it:-)
I guess I will need to spend a little more time thinking about this, and, once I solve it, I will make sure to post the answer.

Thanks for trying to help me with the problem!
• Jun 16th 2013, 06:19 AM
mnov
Re: Implicit differentiation formula
Quote:

Originally Posted by norton429

Here is why I think I am missing something:
I was trying to solve the following problem: find the first partial derivatives of U and V with respect to X and Y and the first partial derivatives of X and Y with respect to U and V, given 2U - V + X^2 + X*Y = 0, U + 2V + X*Y - Y^2 = 0.

To find dU/dX I substituted the V in the second equation from the first equation: V=2U + X^2 + XY => I get eventually: 5U + 2X^2 + 3X*Y - Y^2 = 0. Now, I find Fx = 4X + 3Y, and Fu = 5, thus: dU/dX = -(4X + 3Y)/5, which is the correct answer. Now, if the inverse was true, I should have dX/dU = - 5/(4X + 3Y) but the correct answer seems to be: (4Y - X)/2(X^2 - 2XY -Y^2), and this is why I think I am missing something.

Thanks again!

The correct answer is
$\frac{\partial u}{ \partial x} = \frac{1}{\frac{\partial x}{ \partial u}} = - \frac{4x+3y}{5}$.

Where did you get $\frac{4y-x}{2 (x^2 - 2 xy -y^2)}$ and why does that seem to you to be be correct answer?
• Jun 16th 2013, 01:59 PM
norton429
Re: Implicit differentiation formula
I finally solved the problem (Happy), and here is the solution:
First, if we need to find the first partial derivatives of x and y with respect to u and v, given 2u - v + x^2 + x*y = 0, u + 2v + x*y - y^2 =0 it is important to be clear that in this case u and v are independent variables and x and y are the dependent ones. This seems obvious, of course, but when I was actually trying to solve the problem, I wasn't correctly considering it. So:
1. Differentiate the first equation with respect to u: 2 + 2x*∂x/∂u +y*∂x/∂u +x*∂y/∂u = 0 2. Differentiate the second equation with respect to u: 1 + y*∂x/∂u +x*∂x/∂u- 2y*∂y/∂u = 0; from here get ∂y/∂u = (1 + y*∂x/∂u)/(2y-x), replace ∂y/∂u in the first equation: 2 + 2x*∂x/∂u +y*∂x/∂u +x*(1 + y*∂x/∂u)/(2y-x) = 0; From here it is easy to see that, indeed, ∂x/∂u = (4y-x)/2(x^2 - 2xy - y^2). It turned out to be not such a hard problem (it always seems to be easy once it is solved (Happy))
• Jun 16th 2013, 03:09 PM
mnov
Re: Implicit differentiation formula
Quote:

Originally Posted by norton429
I finally solved the problem (Happy), and here is the solution:
First, if we need to find the first partial derivatives of x and y with respect to u and v, given 2u - v + x^2 + x*y = 0, u + 2v + x*y - y^2 =0 it is important to be clear that in this case u and v are independent variables and x and y are the dependent ones. This seems obvious, of course, but when I was actually trying to solve the problem, I wasn't correctly considering it. So:
1. Differentiate the first equation with respect to u: 2 + 2x*∂x/∂u +y*∂x/∂u +x*∂y/∂u = 0 2. Differentiate the second equation with respect to u: 1 + y*∂x/∂u +x*∂x/∂u- 2y*∂y/∂u = 0; from here get ∂y/∂u = (1 + y*∂x/∂u)/(2y-x), replace ∂y/∂u in the first equation: 2 + 2x*∂x/∂u +y*∂x/∂u +x*(1 + y*∂x/∂u)/(2y-x) = 0; From here it is easy to see that, indeed, ∂x/∂u = (4y-x)/2(x^2 - 2xy - y^2). It turned out to be not such a hard problem (it always seems to be easy once it is solved (Happy))

So that means $\frac{\partial x}{\partial u} \neq 1/ \frac{\partial u}{\partial x}$? I am confused. Can somebody clarify?
• Jun 16th 2013, 10:02 PM
norton429
Re: Implicit differentiation formula
Yes, as you can tell, I was confused too. The thing is that in this case, it is a different condition and, therefore, the formula doesn't apply: here we have two dependent variables (x and y) not one as in the F(x,y,z)= 0 equation. So, what we have in this case is F(x, y, u, v)= 0 and G(x, y, u, v) = 0. This can be solved either the way I described above, or by using Jacobian method. The reason I was able to solve the du/dx as -Fx/Fu is because when you substitute v in the second equation from the first one, you now have an F(x, y, u) = 0 equation and you can apply the -Fx/Fu formula. When you try to find the inverse, this is no longer the case unless you substitute y in one of the equations for x but the result will not be the inverse because we have a different condition: two equations with two dependent variables.
I hope this clarifies the issue.