Implicit Differentiation

stiofan85

Quick question guys:

if I have the equation of a curve in x and y, to find dy/dx can I just differentiate with respect to x, then differentiate the original equation with respect to y, then invert the second so it will be dy/d and multiply d/dx . dy/d and I get dy/dx

It has worked on some sample questions, but I'm not sure if that's just pot luck. Below is one example:

5x^2 + 5y^2 + 6xy = 0

d/dx f(x,y) = 10x + 6y

d/dy f(x,y) = 10y + 6x

therefore

dy/dx = (10x + 6y) / (-10y - 6x)

the negative is due to inverting the derivative?

Thanks!

LeixlipRed

Multiple conceptual errors there!! I have to go but I'm sure someone will be along to help you shortly.

Eliot Rosewater

When I was in secondary school (I haven't done this in college yet) I just differentiated normally just use the chain rule when it comes to the y.

A quick example:

[latex]\displaystyle f(x,y) = x^{2}-2y^{2} = 0[/latex]

Differentiate

[latex]\displaystyle f(x,y) = \frac{d}{dx}(x^{2})-2\frac{d}{dx}(y^{2})=0[/latex]

What you do is apply the chain rule to the y squared. Consider y is a function of x. When you differentiate y squared you differentiate the whole as normal (y squared becomes 2y) and then multiply by the derivative of the function (in this case dy/dx)

So that:

[latex]\displaystyle f(x,y) = 2x-4y(\frac{dy}{dx})=0[/latex]

Then solve for dy/dx.

stiofan85

@LeixlipRed: i know you know your stuff, so I'll take it that it is wrong-I kinda thought as much. I was just trying to think of a different way of doing it as the person I'm showing doesn't quite "get" the chain rule, so wanted to give an easy way to learn off and just do in the exam.

The gas thing is it works on that example Eliot gave though!

d/dx = 2x

d/dy=-4y

dy/d=1/4y

dy/d x d/dx = 2x/4y

LeixlipRed

You can't calculate [latex]\frac{dy}{dx}[/latex] if it's a function of two variables!! So either it's not a function of two variables and you're trying to calculate the derivative of an implicit function of one variable or you're trying to do something you cannot.

stiofan85

Cool. Thanks for clearing that up.

LeixlipRed

Eliot Rosewater wrote: »

When I was in secondary school (I haven't done this in college yet) I just differentiated normally just use the chain rule when it comes to the y.

A quick example:

[latex]\displaystyle f(x,y) = x^{2}-2y^{2} = 0[/latex]

Differentiate

[latex]\displaystyle f(x,y) = \frac{d}{dx}(x^{2})-2\frac{d}{dx}(y^{2})=0[/latex]

What you do is apply the chain rule to the y squared. Consider y is a function of x. When you differentiate y squared you differentiate the whole as normal (y squared becomes 2y) and then multiply by the derivative of the function (in this case dy/dx)

So that:

[latex]\displaystyle f(x,y) = 2x-4y(\frac{dy}{dx})=0[/latex]

Then solve for dy/dx.

But when you set [latex]f(x,y)=0[/latex] it's no longer a function but a level curve of the function.

The OP needs to clarify what exactly it is he is doing or wants to do.

stiofan85

sorry, you're right it's not a function, it's a curve. I should've been clearer.

Anyways, thanks for clearing that up. I was just trying to think of a different way of doing it.

MathsManiac

What's surprising is the fact that it happens to work in quite a few cases! In light of that, it might be useful to give a counterexample. it doesn't work in this case:

x^3+(x^2)y + y^2 = 0.

Now here's an interesting task:
Determine the class of expressions in x and y for which this completely incorrect method happens to give the right answer!

seandoiler

this "completely wrong" method is in fact correct if we equate to zero

http://en.wikipedia.org/wiki/Implicit_and_explicit_functions#Formula_for_two_variables

maths maniac...your example works using the method of the OP (think you mightn't have checked it)

MathsManiac

Good lordie!

I did check - but wrongly!

When I thought I had a counterexample, I didn't investigate further.

Sincerest apologies to all.

sponsoredwalk

It would be helpful to think of;

[latex] f(x,y) \ = \ 5x^2 \ + \ 5y^2 \ + \ 6xy[/latex]

as;

[latex] z \ = \ 5x^2 \ + \ 5y^2 \ + \ 6xy[/latex]

Just like in calculus where;

[latex] f(x) \ = \ x^2 [/latex]

is the same as;

[latex] y \ = \ x^2 [/latex]

My understanding is that if you set [latex] f(x,y) \ = 0 [/latex] as you would like to do in order to solve for [latex] \frac{dy}{dx} [/latex] is technically legal but what you're doing is determining the activity of the variables x & y only along the line z=0. You're restricting z.

It's like if you consider the z-axis to be the dimension coming out of your sheet of paper, by setting f(x,y) = 0 you're only considering the activity of z on the page, and above z there's "only sky", untapped & lonely...


I may be misunderstanding things too however, If I've confused something let me know