Of One Real Variable – By Pheng Kim Ving
1. Explicitly And Implicitly Defined Functions
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.
The equation y = x2 is of the form y = f(x), and thus explicitly defines a function y of
The equation x2 + y2 = 25 implicitly defines a function y of x
near the point
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).
The x-y equation of this curve implicitly defines a
function y of x near
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,
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
Find the slope of the tangent line to the curve:
at the point (1, 2).
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.
Generally, the finding of dy/dx by implicit differentiation of an x-y equation is carried out as follows:
Differentiate both sides of the equation with respect to x. Regard y as a function of x. Use the chain rule for every
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.
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.
5. Tacit Assumption Of Differentiability
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.
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.
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.
2. Suppose (u – v)/(u + v) = u2/v + 1. Find dv/du in terms of u and v.
Utilizing implicit differentiation we get:
3. Suppose xy = x + y. Find y'' in terms of x and y using implicit differentiation.
xy = x + y,
y + xy' = 1 + y',
4. Prove that if Ax2 + By2 = C, then d2y/dx2 = –AC/B2y3.
5. Find an equation of the tangent line to the curve x/y + ( y/x)3 = 2 at the point (– 1, – 1).
Differentiating the given equation implicitly we get:
Hence, an equation of the tangent line is y = 1(x – (– 1)) – 1, or y = x.
curves are said to be orthogonal at a point of intersection if they have perpendicular tangent lines at
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.
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. It follows that m2 = 4/4 = 1. We have m1m2 = – 1(1) = – 1. Therefore, the
two curves are orthogonal at (2, 4).
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.
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.