Implicit DifferentiationThe functions that we have met so far can be described by expressing one variable explicitly in terms of another variable.

For example, , or y = x sin x, or in general y = f(x).

However, some functions are defined implicitly.

Some examples of implicit functions are:

x2 + y2 = 25

x3 + y3 = 6xy

In some cases, it is possible to solve such an equation for y as an explicit function (or several functions) of x.

The graphs of and (x) are the upper and lower semicircles of the circle x2 + y2 = 25.

Nonetheless, x3 + y3 = 6xy is the equation of a curve called the folium of Descartes shown here and it implicitly defines y as several functions of x.

It’s not easy to solve equation x3 + y3 = 6xy for y explicitly as a function of x by hand.

A computer algebra system has no trouble.

However, the expressions it obtains are very complicated

Fortunately, we don’t need to solve an equation for y in terms of x to find the derivative of y.

Instead, we can use the method of implicit differentiation. This consists of differentiating both sides of the equation with respect to x and then solving the resulting equation for y’. In the examples, it is always assumed that the given equation determines y implicitly as a differentiable function of x so that the method of implicit differentiation can be applied.

a. If x2 + y2 = 25, find .

b. Find an equation of the tangent to the circle x2 + y2 = 25 at the point (3, 4).

a. Differentiate both sides of the equation x2 + y2 = 25:

example:

Chain rule:

𝑑𝑓𝑑𝑥=

𝑑𝑓𝑑𝑦

𝑑𝑦𝑑𝑥 =(2 𝑦)(𝑦 ′)

𝑑𝑓𝑑𝑥=

𝑑𝑓𝑑𝑦

𝑑𝑦𝑑𝑥

At the point (3, 4) we have x = 3 and y = 4. 34

dydx

Thus, an equation of the tangent to the circle at (3, 4) is: y – 4 = – ¾(x – 3) or 3x + 4y = 25.

a. Find y’ if x3 + y3 = 6xy.b. Find the tangent to the folium of Descartes x3 + y3 = 6xy at the point (3, 3).c. At what points in the first quadrant is the tangent line horizontal?

Differentiating both sides of x3 + y3 = 6xy with respect to x, regarding y as a function of x, and using the Chain Rule on y3 and the Product Rule on 6xy, we get:

3x2 + 3y2y’ = 6xy’ + 6y

or x2 + y2y’ = 2xy’ + 2y

Now, we solve for y’:

example:

So, an equation of the tangent to the folium at (3, 3) is:y – 3 = – 1(x – 3) or x + y = 6.

When x = y = 3, y’ = – 1

The tangent line is horizontal if y’ = 0.

Using the expression for y’ from (a), we see that y’ = 0 when 2y – x2 = 0 (provided that y2 – 2x ≠ 0).

Substituting y = ½x2 in the equation of the curve, we get

x3 + (½x2)3 = 6x(½x2)

which simplifies to

x6 = 16x3.

Since x ≠ 0 in the first quadrant, we have x3 = 16.

If x = 161/3 = 24/3, then y = ½(28/3) = 25/3.

Thus, the tangent is horizontal at (0, 0) and at (24/3, 25/3), which is approximately (2.5198, 3.1748).

22 siny x y

22 sind d dy x ydx dx dx

This can’t be solved for y.

2 2 cosdy dyx ydx dx

2 cos 2dy dyy xdx dx

22 cosdy xydx

22 cos

dy xdx y

Find the equations of the lines tangent and normal to the curve at2 2 7x xy y ( 1,2)

2 2 7x xy y

2 2 0dydyx yx ydxdx

Note product rule.

2 2 0dy dyx x y ydx dx

22dy y xy xdx

22

dy y xdx y x

at

We need the slope. Since we can’t solve for y, we use implicit differentiation to solve for y’.

example:

tangent:

42 15

y x

4 425 5

y x

4 145 5

y x

normal:

52 14

y x

5 524 4

y x

5 34 4

y x

2 2 12 2 1

m

2 24 1

45

Higher Order Derivatives

Find if 2

2

d ydx

3 22 3 7x y

3 22 3 7x y

26 6 0x y y

26 6y y x

266xyy

2xyy

2

2

2y x x yyy

2

2

2x xy yy y

2 2

2

2x xyy

xyy

4

3

2x xyy y

Substitute back into the equation.

y

Find y” if x4 + y4 = 16.

Differentiating the equation implicitly with respect to x, we get

4x3 + 4y3y’ = 0.

example:

To find y’’, we differentiate this expression for y’ using the Quotient Rule and remembering that y is a function of x:

However, the values of x and y must satisfy the original equation

x4 + y4 = 16.

So, the answer looks quite simple: