## The Calculus Primer (2011)

### Part III. Differentiation of Algebraic Functions

### Chapter 11. DIFFERENTIATION OF IMPLICIT FUNCTIONS

**3—9. Explicit and Implicit Functions.** Consider the function *y*^{2} = 2*px*. When solved for *y*, we have *y* = or *f*(*x*) = where *y* = *f*(*x*). When solved for *x*, we have *x* = *y*^{2}/2*p*, where *x* = *ϕ*(*y*). The functions *y* = *f*(*x*) and *x* = *ϕ*(*y*) are *explicit* functions. In the first instance, *y* is expressed as a function of *x*; in the latter case, *x* is expressed as a function of *y*. Each function is the inverse of the other. But in the original form *y*^{2} = 2*px*, each variable is defined *implicitly* as a function of the other. Additional examples of implicit functions are given herewith:

*x*^{2} + *y*^{2} + 2 = 0;

*x*^{2} + *xy* + *y*^{2} = 1;

*x*^{3} + *y*^{3} = *kxy*;

*x* + *y* =

Implicit functions, instead of being given as *y* = *f*(*x*), or *x* = *ϕ*(*y*), or *s* = *f*(*t*), are often written as *F*(*x*,*y*) = 0. It may be possible to solve an implicit function *F*(*x*,*y*) = 0 for one of the variables in terms of the other, yielding an explicit function such as *y* = *f*(*x*); sometimes this cannot be done conveniently, and sometimes not at all.

**3—10. Differentiation of Implicit Functions.** Although it is not always simple or possible to obtain an explicit function from a given implicit function, nevertheless the derivative can be found, as shown by the following examples.

EXAMPLE 1.Find in the equation *x*^{2} + *y*^{2} = 25.

*Solution*.Differentiate each side of the equation with respect to *x*:

Solve for :

EXAMPLE 2.Find in the equation *y*^{2} = *x*^{3}.

*Solution*.Differentiating each side:

2*y* = 3*x*^{2}.

Solving for :

EXAMPLE 3.Find in the equation *xy* = *a*^{2}.

*Solution*.

EXAMPLE 4.Find in the equation *x*^{2} + *y*^{2} − 2*xy* = 0.

*Solution*.2*x* + 2*y* − 2*x* − 2*y* = 0.

(2*y* − 2*x*) = 2*y* − 2*x*,

= 1.

EXAMPLE 5.Find in the equation *x*^{3} − 2*x*^{2}*y* + 2*y*^{3} = 0.

*Solution*.3*x*^{2} − 2*x*^{2} − 4*xy* + 6*y*^{2} = 0.

(6*y*^{2} − 2*x*^{2}) = 4*xy* − 3*x*^{2},

It should be observed that, in general, as in all the above examples except Example 4, the value obtained for contains both *x* and *y*. If we wish to obtain an expression for the derivative containing only *x*-terms, we may theoretically replace *y* by its value in terms of *x* as found from the original equation, *F*(*x*,*y*) = 0; however, this is sometimes very inconvenient, and is usually not necessary.

It is also worth noting that for most implicit functions, it is generally more convenient to find by the method shown above than it is first to express the given function explicitly and then to differentiate directly.

**EXERCISE 3—3**

*Find* *for each of the following, leaving the remit in terms of x and y:*

**1.** *x*^{2} + *y*^{2} = 36

**2.** *x*^{2} = 4*py*

**3.** *xy* = − 12

**4.** *x*^{2} − *y*^{2} + *x* − *y* = 0

**5.** *x*^{2} + *y*^{2} + *xy* = 0

**6.** 2*x* + *y* =

**7.** *x*^{3}(*x* + *a*) = *y*^{2}

**8.** *x* + *y* = 2*xy*

**9.** *b*^{2}*x*^{2} + *a*^{2}*y*^{2} = *a*^{2}*b*^{2}

**10.** *b*^{2}*x*^{2} − *a*^{2}*y*^{2} = 1

**11.** *xy*^{2} = 3(*x* + 2)

**12.** *x*^{2} − *y*^{2} + *xy* + 2*y* = 4

**13.** *x* − *y* + = *k*

**14.**

**15.** *x*^{3} + *y*^{3} = 3*axy*

**16.** *x*^{3} + 2*x*^{2}*y* + 3*y*^{3} = 0

**EXERCISE 3—6**

**Review**

*Differentiate:*

*Find for each of the following:*

**13.** *x*^{2} + *y*^{2} + 2*x* − 4*y* = 0

**14.** *x*^{3} − *xy* + *y*^{2} = 1

**15.** *x*^{2} + *y* − *x*^{2}*y* = 4

**16.** *x*^{4} + *x*^{2}*y*^{2} + *y*^{4} = 16

**17.** *a*^{2}*x*^{2} − *b*^{2}*y*^{2} = *a*^{2}*b*^{2}

**18.** *x*^{2}*y*^{2} = 2(*x*^{2} − *y*^{2})

**19.** *Ax*^{2} + *Cy*^{2} + *Dx* + *Ey* = 0

**20.** *x*^{⅔} + *y*^{⅔} = *a*^{⅔}