Lesson 2.5: Implicit Differentiation

Graph of 2x^2+3y^2=4
Graph of 2x^2+3y^2=4

We now generalize our definition of a function of one variable to entities that are not truly functions at all, but rather relations. However poorly the terminology has been chosen, relations expressed in such forms as 2x^2+3y^2=4 (an ellipse, with the graph shown to the right) are called implicit functions.

Let’s return to our usual challenge: finding the slope of the (implicit) curve at an arbitrary point (x,y). In this case, we can solve the implicit relation for y as an explicit function of x; we get y=(plus or minus)root((2/3)(2-x^2)). Now we could split up the function into components: one for the positive radicals and one for the negative; apply various derivative rules, and thus find dy/dx = (plus or minus)(x root 6)/(3 root(2-x^2)). That’s an ugly procedure and an even uglier result.

We can use the chain rule as an easier way to find the derivative. Recall the Leibnizian form of the chain rule: df/dx = df/xy * dy/dx. We attack our implicit function term-by-term: differentiating 2x^2 with respect to x is easy; d/dx(2x^2)=4x. Now we have what we might call f(y)=3y^2. We want to differentiate f with respect to x, but we only have f as a function of y. However, the chain rule tells us we can differentiate f with respect to y and then multiply by dy/dx to get df/dx. So we have d/dy(3y^2)=6y, and we multiply by dy/dx to get 6y(dy/dx). Finally, we move to the other side of the equation, where d/dx(4)=0.

Thus our derivative is 4x+6y(dy/dx)=0. We want to solve this for dy/dx, so dy/dx=-(2x)/(3y).

This is a much prettier form than dy/dx = (plus or minus)(x root 6)/(3 root(2-x^2)), but some tedious algebra along with recalling that y=(plus or minus)root((2/3)(2-x^2)) will show that the two expressions are equivalent.

Graph of y^3=x+y
Graph of y^3=x+y

Implicit differentiation, as this process is called, is not only useful for finding derivatives in a more convenient form. There are many circumstances in which we can’t solve an implicit relation for y(x) explicitly. For example, y^3=x+y cannot be solved for y generally. However, we can differentiate the left side with respect to y, finding d/dy(y^3)=3y^2, and then multiply by dy/dx for d/dx(y^3)=3y^2(dy/dx). The first term on the right has d/dx(x)=1, and the second term is differentiated by d/dy(y)=1, so d/dx(y)=1(dy/dx). Therefore our differentiated relation is 3y^2(dy/dx)=1+dy/dx. We subtract dy/dx from both sides and factor the left, so dy/dx(3y^1-1)=1, or dy/dx=1/(3y^2-1).

The TI-89’s CAS can perform implicit differentiation. We do need to give it an expression that is equal to zero, so if we wish to find d/dx(y^3=x+y), we enter d(y(x)^3-x-y(x),x). It’s important to enter y(x) instead of just y, to make the calculator regard y as a function of x rather than as a constant.

The calculator will then return (3y(x)^2-1)(dy/dx)-1, which is equal to zero and so easily solved for dy/dx to give the exact same answer we had before.

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