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Implicit Differentiation

When one writes a formula for a function, say f(x)=3x2+5x+1, one is said to have defined the function explicitly. Similarly, when one writes y=3x2+5x+1, we have explicitly defined y in terms of x.

However, for some equations relating x and y, like x62x=3y6+y5y2, there is no easy way to solve for y explicitly in terms of x. Still, there may exist functions f(x) so that if we plug y=f(x) into this equation, it holds true. The possible functions f(x) that have this property are then said to be defined implicitly by the given equation.

With the assumption that a given equation implicitly defines at least one differentiable function, one can find the derivative of one variable in terms of another through a process known as implicit differentiation.

Consider the following:


Suppose we are asked to find dydx given

3x4y27xy3=48y

Our strategy is to treat the left and right sides of the equation above as functions of x in their own right, and differentiate them both with respect to x. As the left and right sides agree, their derivatives should be equal as well. This will give us an equation in terms of an unknown dy/dx for which we may then solve.

Importantly, as we differentiate each side, we treat y as a function of x. This requires that whenever we take a derivative of a function of y (e.g., y2, or y3), we must invoke the chain rule and pick up an additional factor of dy/dx.

Thus, as an example, just as ddx([f(x)]3)=3[f(x)]2f(x), we find ddx(y3)=3y2dydx.

Differentiating in this way both sides of the equation above results in the following (after applying the product rule twice):

[12x3y2+3x42ydydx][7y3+7x3y2dydx]=08dydx

Note how the result is linear in terms of the dy/dx. Nicely, if the original equation is in terms of only xs and ys, this always happens! As such, we can always easily solve for this unknown dy/dx, by collecting all of the dy/dx terms on one side, with all other terms on the other. Doing this here produces the following:

(6x4y21xy2+8)dydx=7y312x3y2

Dividing by the factor to the left of dy/dx, we then have

dydx=7y312x3y26x4y21xy2+8

As one can see, the resulting expression for dy/dx does involve y values. This almost always happens -- but it should be no surprise, as the original equation we were given may actually implicitly define multiple functions.

To understand this better, suppose we started instead with x2+y2=25, whose graph is a circle. The two functions f1 and f2 that are implicitly defined by this equation correspond to the semi-circles on the top and bottom halves of this circle, as shown below. For any given x value (e.g., x=3), we thus have two possible tangent lines with two potentially different slopes (and hence, values of dy/dx). To find the slope of a tangent line to the curve at some x-value, we need to know to which function we hope to be tangent. The y-value gives us this information.