Implicit Differentiation

 

1.  Explicitly And Implicitly Defined Functions

Consider the equation y = x². Clearly each value of x is mapped to exactly one value of y. So the equation y = x² defines a function y of x. Graphically, a vertical line can meet the curve y = x² at at most one point; see Fig. 1. The equation y = x² gives y explicitly in term of x. It explicitly defines y as a function of x. In general, we say that the equation of the form y = f(x) explicitly defines a function y of x.

 

Generally, let a plane curve be given by an x-y equation. The curve may or may not be the graph of a function. A piece of it extending a short distance on both sides of a point (x₁, y₁) on it may or may not be the graph of a function. In Fig. 2, the curve is not the graph of a function, a piece of it near the point (3, 4) is, any piece of it near (extending on both sides of) the point (5, 0) is not. In Fig. 3, the curve is not the graph of a function, but a piece of it near the point (x₁, y₁) is. Thus, as in Fig. 3, we say that the x-y equation implicitly defines a function y of x near the point (x₁, y₁).

2.  Implicit Differentiation



In this second method of finding the slope, we differentiate the equation x² + y² = 25 itself with respect to x, regarding y as a function of x. This is the differentiation of an equation which implicitly defines a function y of x. It's thus called implicit differentiation. Observe that we differentiate both sides of the equation, and that since y is a function of x, the chain rule must be used for every y-term.

Let's find dy/dx where x and y are linked by our hard-to-solve-for-y equation x⁶ + 3y⁵ – y² + x²y = 2
using implicit differentiation. We have 6x⁵ + 15y⁴(dy/dx) – 2y(dy/dx) + 2xy + x²(dy/dx) = 0.
Therefore, dy/dx = (– 6x⁵ – 2xy)/(15y⁴ – 2y + x²).



3.  Dependence Of dy/dx On Both x And y

The derivative dy/dx = 2x of y = x² depends only on x. Now, the derivative or slope dy/dx = – x/y in the circle example depends on both x and y. Examine the slope of the tangent line to the circle at x = 3. Clearly there are two tangent lines at x = 3. One is to the upper semi-circle at the point (3, 4), and the other is to the lower semi-circle at the point (3, – 4). The points (3, 4) and (3, – 4) are different but have the same x-coordinate of 3. So, to select one among them, it's not enough to specify the x-coordinate alone; we must specify the y-coordinate too. Hence, to find the slope at a point, both coordinates of that point must be known. That's why the derivative also depends on y
 
This phenomenon can occur in general. This is because an x-y equation can define more than one functions y of x, and consequently the choice as to the derivative of which one function must be made.




4.  Solving For dy/dx

In the implicit differentiation of an x-y equation, the chain rule is used with every y-term (eg, y³ or sin y). The derivative of such a term thus ends up in dy/dx (eg, (d/dx) y = dy/dx, (d/dx) x²y³ = 2xy³ + 3x²y²(dy/dx), using (d/dx) sin x = cos x: (d/dx) 2 sin y = 2 cos y (dy/dx)). After the implicit  differentiation, we solve for dy/dx. This is easy, because dy/dx always appears in the first power only.



5.  Substituting The Coordinates Of The Point Before Solving For dy/dx

Example 1

Find the slope of the tangent line to the curve:



After differentiation, we can solve for dy/dx in terms of x and y before substituting x = 1 and y = 2 to get (dy/dx)|_{x=1, y=2}. But we don't. Instead, we substitute x = 1 and y = 2 before solving for dy/dx to obtain (dy/dx)|_{x=1, y=2}.

In general, we substitute the coordinates of the given point as soon as we differentiate the given equation, and then solve the resulting equation for the derivative. With numbers substituted for x and y, it's usually much easier to solve for the derivative than it would be with algebraic symbols x and y.



6.  Carrying Out The Finding Of dy/dx By The Implicit
     Differentiation

Generally, the finding of dy/dx by the implicit differentiation of an x-y equation is carried out as follows:

i.  Differentiate both sides of the equation with respect to x. Regard y as a function of x. Use the chain rule for every y-term.

ii.  If you need the derivative corresponding to some point (x₁,y₁) on the curve, substitute x = x₁ and y = y₁.

iii.  Solve for dy/dx.



7.  Finding Higher-Order Derivatives

For example, let's find y'' if x² + y² = 25. Differentiating this equation implicitly we get 2x + 2yy' = 0.
So y' = – x/y. Differentiating this equation we obtain y'' = (– y + xy')/y² = (– y + x(– x/y))/y² = – (x² + y²)/y³. Note that there's no y' in the expression for y''. In the expression for y'', we must replace y' with its expression in x and y.



8.  Caution: Tacit Assumption Of Differentiability

When we implicitly differentiate an x-y equation, we tacitly assume that whatever function implicitly defined by the equation is differentiable. This assumption is implied, for otherwise we would attempt to calculate a quantity (the derivative) that may or may not exist.



Problems & Solutions


1.  Suppose y² – 2xy + 3x² = 1 and (x₁, y₁) = (0, – 1). Use each of the following methods to find dy/dx when x = x₁ for y defined implicitly as a function of x near (x₁, y₁).
a.  Find y explicitly as a function of x and differentiate.
b.  Perform implicit differentiation.

Solution





2.  Suppose (u – v)/(u + v) = u²/v + 1. Find dv/du in terms of u and v.

Solution

Utilizing implicit differentiation we get:




3.  Suppose xy = x + y. Find y'' in terms of x and y using implicit differentiation.

Solution

xy = x + y,
y + xy' = 1 + y',

4.  Prove that if Ax² + By² = C, then d²y/dx² = –AC/B²y³.

Solution





5.  Find an equation of the tangent line to the curve x/y + ( y/x)³ = 2 at the point (– 1, – 1).

Solution





6.  Two curves are said to be orthogonal  at a point of intersection if they have perpendicular tangent lines at that point.
a.  Prove that the curves 2x² + y² = 24 and y² = 8x are orthogonal at the point (2, 4) of intersection.
b.  Prove that for any value of c and any non-zero value of k, the curves y² – x² = c and xy = k are orthogonal at all points of intersection.

Solution

a.  Let m₁ be the slope of the tangent line to the curve 2x² + y² = 24 and m₂ that to the curve y² = 8x at (2, 4). Implicit differentiation of 2x² + y² = 24 gives 4x + 2y y' = 0, so y' = – 2x/y. Thus, m₁ = – 2(2)/4 = – 1. Implicit differentiation of y² = 8x yields 2y y' = 8, hence y' = 4/y. Therefore, m₂ = 4/4 = 1. We have m₁m₂ = – 1(1) = – 1. Consequently, the two curves are orthogonal at (2, 4).

b.  Let (u, v) be an arbitrary point of intersection, m the slope of the tangent line to the curve y² – x² = c at (u, v), and n that to the curve xy = k at (u, v). Differentiating y² – x² = c implicitly we get 2y y' – 2x = 0, so that y' = x/y. Thus, m = u/v. Differentiating xy = k implicitly we obtain y + xy' =0, so that y' = – y/x. Hence, n = – v/u. Now, (u, v) lies on the curve xy = k; this shows that uv =k ¹ 0; consequently, u ¹ 0 and v ¹ 0. Therefore, both m and n exist. We have mn = (u/v)(– v/u) = – 1. It follows that the two curves are orthogonal at (u, v).



7.  Suppose (x – y)/(x + y) = x/y + 1.
a.  Use implicit differentiation to find dy/dx.
b.  Now prove that the given equation doesn't define any differentiable function y of x. This shows that the derivative calculated above doesn't exist and thus is meaningless.

Solution