Calculus Of One Real Variable By Pheng Kim Ving
Chapter 4: More On The Derivative Section 4.2: Implicit Differentiation

 

4.2
Implicit Differentiation

 

 

Return To Contents
Go To Problems & Solutions

 

1. Explicitly And Implicitly Defined Functions

 

Consider the equation y = x2. See Fig. 1.1. 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. 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.

 

 

Fig. 1.1

 

The equation y = x2 is of the form y = f(x), and thus explicitly defines a function y of
x.

 

Fig. 1.2

 

The equation x2 + y2 = 25 implicitly defines a function y of x near the point
(3, 4).

 

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. 1.2,
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. 1.3, the curve is not the graph of a function, but a piece of it near the point (x1, y1) is.
Therefore, as in Fig. 1.3, we say that the
x-y equation implicitly defines a function y of x near the point (x1, y1).

 

Fig. 1.3

 

The x-y equation of this curve implicitly defines a function y of x near
the point (
x1, y1).

 

Go To Problems & Solutions Return To Top Of Page

 

2. Implicit Differentiation

 

Let's find the slope of the tangent line to the circle x2 + y2 = 25 at the point (3, 4). Refer to Fig. 1.2. Solving x2 + y2 = 25

 

x6 + 3y5 y2 + x2y = 2.

 

Well, it's pretty hard to solve this one for y. This situation motivates us to seaech for a different method to do the
differentiation without having to first solve the given equation for
y.

 

As an example, let's return to our circle. We want to find the same slope directly from the original equation x2 + y2 = 25
itself, without having to first solve it for
y to get a solution of the form y = f(x). We know that it (implicitly) defines a
function
y of x near the point (3, 4). Hence, in it, y is a function of x near the point (3, 4). It follows that we can talk
about the derivative of
y with respect to x at that point, that derivative being our slope. Now that we can talk about the
derivative
dy/dx, let's find it. In the equation x2 + y2 = 25, the quantities x2 and 25, like y, are also functions of x
(everywhere, in particular near
x = 3) (25 is a constant function). Since x2 + y2 and 25 are equal, their derivatives with
respect to
x are equal too, ie, (d/dx)(x2 + y2) = (d/dx)(25) = 0. Then:

 

 

which is the same as the one obtained by the first method.

 

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 that 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.

 

Now 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, as follows:

 

x6 + 3y5 y2 + x2y = 2,

 

 

Go To Problems & Solutions Return To Top Of Page

 

3. More On Finding dy/dx By Implicit Differentiation

 

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. Consequently, 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

 

 

Solving For dy/dx

 

In the implicit differentiation of an x-y equation, the chain rule is used with every y-term (eg, y or y3 or sin y). It follows
that the derivative of such a term ends up in
dy/dx. For example, (d/dx) y = dy/dx, (d/dx) x2y3 = 2xy3 + 3x2y2(dy/dx),
(
d/dx) 2 sin y = 2 cos y (dy/dx)) (using the formula (d/dx) sin x = cos x, which we'll see later on in this tutorial). After
the implicit differentiation, we solve for
dy/dx. This is easy, because dy/dx always appears in the first power only.

 

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

 

Example 3.1

 

Find the slope of the tangent line to the curve:

 

 

at the point (1, 2).

 

Solution

EOS

 

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 like
x and y.

 

Recap

 

Generally, the finding of dy/dx by 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.

 

Go To Problems & Solutions Return To Top Of Page

 

4. 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:

 

 

Note that there's no y' in the expression for y''. In the expression for y'', we must replace y' by its expression in x and y.

 

Go To Problems & Solutions Return To Top Of Page

 

5. 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.

 

This assumption is valid for all of the above examples. However, it's not always valid, as demonstrated in Problem &
Solution 7
. In multi-variable calculus, there's a theorem that gives the conditions under which an implicitly-defined
function is differentiable.

 

Return To Top Of Page

 

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

 

 

 

Return To Top Of Page

 

 

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

 

Solution

 

Utilizing implicit differentiation we get:

 

 

Return To Top Of Page

 

 

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

 

Solution

 

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

 

 

Return To Top Of Page

 

 

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

 

Solution

 

 

Return To Top Of Page

 

 

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

 

Solution

 

Differentiating the given equation implicitly we get:

 

 

Hence, an equation of the tangent line is y = 1(x ( 1)) 1, or y = x.

 

Return To Top Of Page

 

 

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 2
x2 + 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. It follows that m2 = 4/4 = 1. We have m1m2 = 1(1) = 1. Therefore, the
two curves are orthogonal at (2, 4).

 

 

Return To Top Of Page

 

 

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

 

 

 

Note

 

Now we see that we can't be sure that the differentiability assumption is always valid; see Part 5. An x-y equation
doesn't always implicitly define a differentiable function
y of x.

 

Return To Top Of Page Return To Contents