Implicit Differentiation Sometimes we have a curve in the plane defined by a function Fxy= (, ) 0 (1) that involves both x and y, possibly in complicated ways. Examples. We show below the graphs of the two plane curves that are the solution points to the equations y3 y+ y x = 0 and ( x 3) + ( y 3) 8= 0
One problem we might want to solve is the problem of finding the equation of the tangent line to these curves at some point. If the given curve is the graph of a function f(x), and if we can solve the equation F(x, y) = 0 for y, we will have the formula f(x), and we can take the derivative of that function with respect to x, to compute the slope of the tangent line to the curve at a given point on the curve. Two problems present themselves: Problem 1. The curve may be the graph of a function, but it could be impossible or impractical for us to solve the equation F(x, y) = 0 for y.
Example. Look again at the curve y3 y+ y x = 0 whose graph is shown below. Clearly this curve satisfies the vertical line test and so is the graph of a function f(x). We cannot solve the equation y3 y+ y x = 0for y in terms of x, by any reasonable method, so we cannot find the formula f(x) for this function. Thus we cannot find f (x), and therefore the equation of the tangent line at a point, in any direct way.
Problem. The curve might not satisfy the vertical line test, so that it is not the graph of a function. Then to stu the tangent line at a point, we might be able to view part of the curve that includes that point as the graph of a function. Example. Look at the curve ( x 3) + ( y 3) 8= 0 That we viewed before. This is the circle shown below (3, 3)
This circle does not satisfy the vertical line test and so is not the graph of a function. However, the upper and lower halves are the respective graphs of the functions 3+ 8 ( x 3) found by solving for y. and 3 8 ( x 3) y= 3+ 8 ( x 3) (3, 3) y= 3 8 ( x 3)
To find the slope of tangent line at the red point, we could use the upper function, and take the derivative at that point. To find the slope of the tangent line at the blue point, we could take the derivative of the bottom function at that point. y= 3+ 8 ( x 3) (3, 3) y= 3 8 ( x 3)
This procedure is quite complicated, and may be much more so for more difficult formulas. Also, as for problem 1, it may be a practical impossibility to solve for the piece of the curve you need in the form y = f(x). Definition. We say that an equation of the form F(x, y) = 0 defines y as a function of x implicitly around the point (x 0, y 0 ) if a piece of the curve F(x, y) = 0 containing (x 0, y 0 ) satisfies the vertical line test.
This is illustrated below. The part of the curve shaded blue on the right is the graph of some function f(x), and f (x 0 ) is the slope of the tangent line shown in green. However, the formula f(x) will vary with the point chosen, and it is likely to be very difficult or impossible to solve for this formula and take its derivative. (x 0, y 0 ) (x 0, y 0 ) y = f(x) F(x, y) = 0 F(x, y) = 0
There is, however, a way of finding the number f (x 0 ) without having the formula for f (x). It is called implicit differentiation. We illustrate it in the following example. Example. Consider the curve x 3 y 4 1 = 0. Find the slope of the tangent line to this curve at the point (1, 1). Implicit Solution: Assume that y is defined implicitly as an (unknown) function f(x) of x near x = 1, so that f(1) = 1. Take the derivative of both sides of the equation and get: 3x y4 x3 4y 3 + = 0
Here we used the product formula, and on the term y 4 we used the chain rule, since we are assuming that y is a function of x. We then solve for the derivative to get 4 = 3x y 4x 3 y 3 At the point (1, 1) we see that 3 = 4 Thus the tangent line to the curve x 3 y 4 = 1. at the point (1, 1) has the equation ( y 1) = 3 ( x 1) 4 7 3 or y= x 4 4
We summarize this process as follows: Start with the equation F(x, y) = 0. If we assume that y is defined implicitly as a function of x, y = f(x), then the entire left hand side of the equation, F(x, y), is really a function of x, and we can take its x-derivative. To get the x-derivative of the part that involves y, we use the chain rule. When we are done, we set the result equal to the derivative of the right hand side, which is zero. We solve this result for the unknown quantity
Example. Let us look again at the circle defined by ( x 3) + ( y 3) 8= 0 or ( x 3) + ( y 3) = 8 Find the slope of the tangent line at the point (5, 5). Hard Way: The equation of the function whose graph contains (5, 5) (the red graph) is f() x = 3+ 8 ( x 3) (5, 5) (3, 3)
We take the derivative of f() x = 3+ 8 ( x 3) getting [8 ( 3) f () x = x ] = [ ( x 3)] = [( x 3)] 8 ( x 3) 4 ( x 3) 8 ( x 3) f (5) [ ] So the slope of the tangent line at (5, 5) is = = = 1 8 () This is shown below (5, 5) (3, 3)
Easy Way: take the derivative of both sides of the equation ( x 3) + ( y 3) 8= 0 treating y as a function defined implicitly, and using the chain rule. Then solve for the unknown derivative. ( 3) ( 3) ( 3) ( 3) x + y = 0. Thus = x = x ( y 3) ( y 3) And at (5, 5) this is 1.
Example: Find the slope of the tangent line to the curve xy = 1 at the point (, 1/ ). Usual solution: This curve is the graph of the function 1 The derivative of this function is x Thus the slope of the tangent line at the given point is 1 f() x = x 1 4 Implicit Solution: Differentiating the original equation, we get y+ x = 0, or = y = 1 x 4
Example: Show that the formula [ r ] = rxr 1 Holds for all rational numbers r = p q Solution: If p 1 y = xr = xq = x p q then y q = x p. Differentiating both sides, we have q 1 p 1 qy = px
so qx ( 1) pq q = px p 1 or qx p p q p 1 = px Solving for the derivative, we get. p ( p 1) p+ p q 1 = p x = p xq = rxr 1 q q
Example. Use the previous formula to find the derivative of 3 x 1 f() x = x+ Solution. By the chain rule and the previous formula, the derivative is 1 1 3 x 1 x 1 3 x 1 ( x+ ) ( x 1) = x+ x+ x+ ( x ) + 1 3 x 1 = 3 x + ( x+ )
Example. Use implicit differentiation to find if xy 3 5xy+ x= 1 Solution. 3xy+ x3y 10xy 5x += 1 0 xy 3 5x = 3xy+ 10xy 1 x 3 y 5x 3xy = + 10xy 1 3x y 10xy 1 = + x 3 y 5 x
Example. Use implicit differentiation to find if tan( xy) xy= 0 Solution. sec ( ) xy y x y x + = 0 sec ( ) sec y xy + x ( xy) y x = 0 sec x ( xy) x ysec = ( xy) + y ysec ( xy) y = + x sec ( xy ) x
Example. Use implicit differentiation to find if x3+ y3= 1 Solution. 3 3 x + y = 0 = x y () y x + x y d y = y4 d y y () x + x y d y = y4 x y x4 xy = y y4