Mastering Calculus - Implicit Differentiation

Of the two branches of calculus, integral and differential, the latter admits to procedure while the former admits to creativity. This notwithstanding, the realm of implicit differentiation provides substantial room for confusion, and this topic often hinders a student's progress in the calculus. Here we look at this procedure and clarify its most stubborn features.

Normally when differentiating, we are given a function y defined explicitly in terms of x. Thus the functions y = 3x + 3 or y = 3x^2 + 4x + 4 are two in which the dependent variable y is defined explicitly in terms of the independent variable x. To obtain the derivatives y', we would simply apply our standard rules of differentiation to obtain 3 for the first function and 6x + 4 for the second.

Unfortunately, sometimes life is not that easy. Such is the case with functions. There are certain situations in which the function f(x) = y is not explicitly expressed in terms of the independent variable alone, but is rather expressed in terms of the dependent one as well. In some of these cases, the function can be solved so as to express y solely in terms of x, but often times this is impossible. The latter might occur, for example, when the dependent variable is expressed in terms of powers such as 3y^5 + x^3 = 3y - 4. Here, try as you might, you will not be able to express the variable y explicitly in terms of x.

Fortunately, we can still differentiate in such cases, although in order to do so, we need to admit the assumption that y is a differentiable function of x. With this assumption in place, we go ahead and differentiate as normal, using the chain rule whenever we encounter a y variable. That is to say, we differentiate any y variable terms as though they were x variables, applying the standard differentiating procedures, and then affix a y' to the derived expression. Let us make this procedure clear by applying it to the above example, that is 3y^5 + x^3 = 3y - 4.

Here we would get (15y^4)y' + 3x^2 = 3y'. Collecting terms involving y' to one side of the equation yields 3x^2 = 3y' - (15y^4)y'. Factoring out y' on the right hand side gives 3x^2 = y'(3 - 15y^4). Finally, dividing to solve for y', we have y' = (3x^2)/(3 - 15y^4).

The key to this procedure is to remember that every time we differentiate an expression involving y, we must affix y' to the result. Let us look at the hyperbola xy = 1. In this case, we can solve for y explicitly to obtain y = 1/x. Differentiating this last expression using the quotient rule would yield y' = -1/(x^2). Let us do this example using implicit differentiation and show how we end up with the same result. Remember we must use the product rule to xy and do not forget to affix y', when differentiating the y term. Thus we have (differentiating x first) y + xy' = 0. Solving for y', we have y' = -y/x. Recalling that y = 1/x and substituting, we obtain the same result as by explicit differentiation, namely that y' = -1/(x^2).

Implicit differentiation, therefore, need not be a bugbear in the calculus student's portfolio. Just remember to admit the assumption that y is a differentiable function of x and begin to apply the normal procedures of differentiation to both the x and y terms. As you encounter a y term, simply affix y'. Isolate terms involving y' and then solve. Voila, implicit differentiation.