Wikipedia:Reference desk/Archives/Mathematics/2006 November 18

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November 18

Implicit differentiation

Hello. I don't understand how to differentiate equations like $y=sin(x+y)$ for y where y is also part of another function. Could somebody walk me through the steps of this example or a similar one?--El aprendelenguas 23:17, 18 November 2006 (UTC)

Not sure about in general but in this case:

y=sin(x+y) so arcsin (y) = x+y, then x= arcsin (y) - y.

dx/dy can be easily found and by extension dy/dx .

dx/dy = 1/sqrt(1-y*y) - 1 = [1-sqrt(1-y*y)] / sqrt (1-y*y)

dy/dx = sqrt (1-y*y) / (1-sqrt(1-y*y))

In this case I've simply rearranged so that x and y are separate - give another example if you wish..

Probably not what you where asking for? Need more explanation etc or is that ok? —The preceding unsigned comment was added by 87.102.9.39 (talk • contribs) 00:18, November 19, 2006 (UTC).

In the above "sqrt(1−y²)" needs to be replaced by "± sqrt(1−y²)", since x = sin Z has other solutions than Z = arcsin x, such as Z = π − arcsin x.  --Lambiam^Talk 00:42, 19 November 2006 (UTC)

Here is another way:

y = sin(x+y), so (d/dx)y = (d/dx)sin(x+y).

Using the chain rule, this gives:

y' = (1+y')cos(x+y).

The trigonometric function can be eliminated with the identity cos² α + sin² α = 1:

y' = ±(1+y')sqrt(1−y²),

from which you can retrieve the earlier solution by solving for y'. The sign can flip only when |y| = 1, and this differential equation has more solutions (for example, the constant function y = 1) than the original equation. The whole thing is a bit tricky, since you can see when you graph the function that y' → ∞ as x → 0.  --Lambiam^Talk 01:07, 19 November 2006 (UTC)

Taking derivative means to find the rate of change AT A GIVEN POINT (a,b) satisfying $b=sin(a+b)$. Why $y'=cos(a+b)/[1-cos(a+b)]$ is NOT an acceptable answer? Twma 08:09, 19 November 2006 (UTC)

Is it because any values of a and b can be used regardless of whether those values satisfy the original equation? (bit confused, feels like I am back at school ??) —The preceding unsigned comment was added by 83.100.138.99 (talk • contribs) 17:14, November 19, 2006 (UTC).

To me it is, except I used x and y instead of your a and b. Then your answer is essentially equivalent to my y' = (1+y')cos(x+y). I added the next steps to relate this to 87.102.9.39's answer.  --Lambiam^Talk 18:45, 19 November 2006 (UTC)

Since we have an implicit equation, it may be of interest to take a different approach. We are looking at the zero set of the function f(x,y) = y−sin(x+y). A normal to such a curve at a point can be found as the gradient of f, which is the vector ∇f(x,y) = (−cos(x+y),1−cos(x+y)). Rotating the normal clockwise 90° gives a vector tangent to the curve, in this case (1−cos(x+y),cos(x+y)). The slope of this is the derivative of y with respect to x, namely ^cos(x+y)⁄_1−cos(x+y). For example, at (0,0) the curve is vertical (y′ = +∞); at (^π⁄₂−1,1) it is horizontal (y′ = 0); and at (π,0) it tilts downward (y′ = −¹⁄₂). This approach seems both simpler and more flexible.

However, I wonder if the question was really meant to be about how to solve a differential equation, y′ = sin(x+y). --KSmrq^T 09:42, 19 November 2006 (UTC)

I can't make sense of what users "Lambian" and "KSmrq" have done here - can anyone confirm that their maths makes no sense (or otherwise)? (Please read original question) —The preceding unsigned comment was added by 83.100.138.99 (talk • contribs) 12:47, November 19, 2006 (UTC.

Makes sense to me. If you want to find dy/dx for the curve defined by the equation y=sin(x+y) then implicit differentiation gives you dy/dx=cos(x+y)/(1-cos(x+y)) and a little more work gives you dy/dx as a function of y alone, if you want it in that form. Try sketching the curve to see what is happening - probably easiest to use the inverse formula x=arcsin(y)-y, and take values of y from -1 to 1. Gandalf61 13:09, 19 November 2006 (UTC)

Lambian is basically just differentiation both sides of the equation wrt x. When you find an x, you just get dx/dx = 1, when you find a y just replace it by y'=dy/dx, the rest follows from the standard rules of differentiation, in particular the chain rule. This is a purely formal process, applying standard rules, so don't worry too much about the implications. In this example a lot of the complication comes from the particular example chosen, which takes some algebra to turn the formal solution y'=(1+y') cos(x+y) into a solution involving just y'. You may find it easier to consider a simpler equation say y²=x⁴. --Salix alba (talk) 13:35, 19 November 2006 (UTC)

Assuming you're the original questioner, can you say something about your background knowledge? For example, do you know and understand the chain rule? Have you read the article on implicit functions? Can you point out where it is you get stuck when trying to understand it?  --Lambiam^Talk 15:56, 19 November 2006 (UTC)

It was me who said "I can't make sense of..." - I couldn't understand what you two were doing - I added the first reply - gandalf and salix alba explained what you were up to and now it makes a bit more sense.. (No idea what happened to the original question asker)

Say if I needed to differentiate x^3+x^2+2=y^7+y+4 I would use : z=x^3+x^2+2 and find dz/dx and dz/dy. dy/dx is given by (dz/dx)/(dz/dy) That's (effectivly) the same method as you used, yes? —The preceding unsigned comment was added by 83.100.138.99 (talk • contribs) 16:26, November 19, 2006 (UTC).

Yes, if you use z=y^7+y+4 to find dz/dy, but note that this only works if you can separate the variables as in f(x)=g(y). For a case like sin(x − sin y) = cos(y + cos x) you need a more general method.  --Lambiam^Talk 18:11, 19 November 2006 (UTC)

Yes, i get (1-cos y*dy/dx)cos(x-siny)=-(dy/dx-sinx)sin(y+cosx) then rearranging..is this right? —The preceding unsigned comment was added by 83.100.138.99 (talk • contribs) 18:24, November 19, 2006 (UTC).

Looks right to me. If you read Implicit function (and understand partial derivatives), you can see how to do this so that you don't need the tedious rearranging (assuming the purpose is to bring dy/dx out). By the way, please sign your contributions using 4 tildes like this: ~~~~. Thank you.  --Lambiam^Talk 18:55, 19 November 2006 (UTC)

Seem like just using the chain rule - I probably was taught this at some point but forgot it because I never needed to use it. Thanks for bringing it up. Who knows maybe one day I will actually really use it. Thanks.83.100.138.99 19:01, 19 November 2006 (UTC)

My answer is correct; feel free to test its results against any answer you trust more. Does it make sense to you; can you understand it? That's a different question! I apologize if it is not clear to a general reader; I was trying to briefly support Twma's hypothesis.

I'll walk slowly through the example of 83.100.138.99, then through the more challenging example of Lambiam, and try to show the idea more clearly.

We are given the equation x³+x²+2 = y⁷+y+4. Consider all pairs of real numbers, (x,y); some will satisfy this equation, but most will not. Those that do satisfy it trace out a curve in the plane. In general that curve may not have a single intersection with a given vertical line. We need that to define y as a function of x. (Consider x² = 1−y², which is a circle.)

The pivotal step is to transform the problem. First, change the original equation to

$x^{3}+x^{2}+2-y^{7}-y-4=0.\,\!$

We have merely subtracted the right-hand side from the left-hand side. Any (x,y) pair that satisfied the original equation will satisfy this one. Now define a function of two variables,

$f(x,y)=x^{3}+x^{2}+2-y^{7}-y-4.\,\!$

The solutions of the original equations are the same as the solutions of f(x,y) = 0. Now think of f as describing the height of a surface above a given point of the plane, z = f(x,y). The surface can have hills and valleys and ridges and so on, but only one height at each (x,y) position. If we consider z = 0 to be "sea level", positive heights will be dry (above water) and negative heights will be wet (below water). Our solution curve is the shore line. Through any given point on that curve we wish to find the line best matching the shore there.

The gradient of f is the vector whose first component is the derivative of f with respect to x, and whose second component is the derivative of f with respect to y, ∇f = (^∂f⁄_∂x,^∂f⁄_∂y). As discussed in our article, that vector points straight in the "uphill" direction. Consequently, the gradient ∇f is perpendicular to the curve, and so perpendicular to the shoreline, at the given point. We may thus rotate the gradient 90° to get a vector pointing along the shore (a tangent vector).

Applied to our current example, we find

$\nabla f={\begin{bmatrix}\partial f/\partial x\\\partial f/\partial y\end{bmatrix}}={\begin{bmatrix}3x^{2}+2x\\-7y^{6}-1\end{bmatrix}}.$

Rotating this 90° gives the vector

${\begin{bmatrix}7y^{6}+1\\3x^{2}+2x\end{bmatrix}}.$

Our original quest to find the derivative is now fulfilled by taking the ratio of the second component (y) to the first (x),

${\frac {dy}{dx}}={\frac {3x^{2}+2x}{7y^{6}+1}}.$

This only makes sense at a point on the curve, but it works even if y is not single-valued in terms of x. (For example, the circle mentioned above gives f(x,y) = x²+y²−1; its gradient is (2x,2y); a tangent is (−2y,2x); and the derivative is −x/y.) It necessarily fails at any point where a curve intersects itself or abruptly changes direction; at such a point the derivative we seek does not exist.

Is it clear why we can find the derivative from the tangent? If not, ask.

Now let's try Lambiam's example, sin(x−sin y) = cos(y+cos x). Our implicit function will be

$f(x,y)=\sin(x-\sin y)-\cos(y+\cos x).\,\!$

Its gradient is

$\nabla f={\begin{bmatrix}\cos(x-\sin y)-\sin(x)\sin(y+\cos x)\\\sin(y+\cos x)-\cos(y)\cos(x-\sin y)\end{bmatrix}}.$

And the derivative of y with respect to x at a point (x,y) is

$-{\frac {\cos(x-\sin y)-\sin(x)\sin(y+\cos x)}{\sin(y+\cos x)-\cos(y)\cos(x-\sin y)}}.$

We are never required to solve for y as a function of x, nor for x as a function of y. So long as we have a point (x,y) that satisfies the equation, we can substitute its coordinates in the formula to get the derivative there. For example, we can easily verify that Lambiam's equation is satisfied by (0,^π⁄₂), with derivative −1, and also by (0,−^π⁄₂), with derivative +1. Similarly, we can see that (1,0) satisfies our first equation, with derivative 5. --KSmrq^T 06:47, 20 November 2006 (UTC)

Thank you to everyone who contributed!--El aprendelenguas 01:18, 20 November 2006 (UTC)