# Implicit Differentiation Calculus Slope of Line Tangent to Circle

“Implicit differentiation” refers to differentiation of an implicitly defined function.

Ordinarily the definition of a function is explicit: the name of the function appears on one side of an equals sign and the operations on the independent variables that define the function appear on the other, for example, f(x)=x^2+3-sin(x). An implicit function definition, however, contains operations on the function itself, for example, the elliptic curve f(x):f(x)2=x^3+ax+b or the circle f(x):f(x)^2+x^2=r^2. (As in Fortran or Matlab, here “x^2” means “x to the second power”.)

An astute reader will notice that both of these are not true functions, in that they are multi-valued. This is not universally true of implicit functions, but it is not uncommon. In order to find the value or the derivative we will have to specify a branch-we’ll get to that.

Sometimes the function is represented by a single symbol. Instead of f(x) it may just be called f, as in f:f^2+x^2=r^2. The key to implicit differentiation is to remember that f is the function, not an independent variable, and to differentiate both sides of the definition with respect to the independent variable accordingly. This will yield an implicit definition for the derivative, which we may sometimes solve to obtain an explicit function.

Let’s take the circle as an example. For notational convenience, let’s specify a differentiation operator D such that D[f(x)]=du/dx, and the shorthand “prime” notation f’=df/dx.

Begin by differentiating both sides:

D[f^2+x^2]=D[r^2].

Now, recognizing that r is a constant, we have

D[f^2+x^2]=0.

Applying the sum rule for differentiation, this becomes

D[f^2]+D[x^2]=0.

And applying the power rule to the second term, we have

D[f^2]+2x=0.

The following is the step that sets implicit differentiation apart: apply the chain rule and differentiate the first term above. We have

2f*D[f]+2x=0

or

2f*f’+2x=0.

Here the asterisk denotes ordinary multiplication.

We may solve this for f’, obtaining

f’=-x/f.

Note that in this case we must know the value of f to know f’. f itself is not truly a function; it is multi-valued and we must specify which branch (negative or positive) is of interest. f’ on the other hand does not appear multi-valued above. However, for a given x, it can take on two values, one for the positive value of f and one for the negative value of f. These correspond to the slopes of the lines tangent to the top and bottom halves of the circle, respectively.

That’s all there is to it: implicit differentiation is just application of the chain rule to an implicit function definition. As long as one keeps track of the difference between partial and total derivatives and uses the appropriate chain rule, this method can also be used to differentiate implicitly defined functions of multiple variables.

In many cases, however, the solution is not as clean as above, even though the procedure is the same. More often than not, the derivative of an implicitly defined function will itself be an implicitly defined function-there will be no way to solve the equation for f’. All of the arithmetical tools available to compute implicit functions works to compute an implicitly defined derivative. In particular, we note that an implicit definition for a derivative found by implicit differentiation will be of the form

u(f,f’,x)=v(f,f’,x)

where u and v are both functions of the function f, its derivative f’, and the independent variable x. Given the value of x and f for which we want to find the derivative f’, the intersection of u and v, both treated as functions of f’ alone, gives f’. Or we can rearrange, obtaining:

u(f,f’,x)-v(f,f’,x)=0

and solve using Newton’s Method or the secant method.