Implicit Differentiation
Article objectives
Introduction
Sometimes it is inconvenient to differentiate a function explicitly because it is not easy to get the function in an explicit form. A function in explicit form is in the form \(y = f(x)\). The first example illustrates this inconvenience.
Example 1: Consider the following equation:
$$e^{2y\ln(5x – 9)} + 4 = 49x^2 + 42x + 13$$
Although it may look really difficult, it is actually fairly easy to write this equation explicitly in terms of x. Watch the following string of algebraic manipulations:
$$e^{2y\ln(5x – 9)} = 49x^2 + 42x + 9 \Rightarrow$$ $$e^{2y\ln(5x – 9)} = (7x + 3)^2 \Rightarrow$$ $$(e^{y\ln(5x – 9)})^2 = (7x + 3)^2 \Rightarrow$$ $$e^{y\ln(5x – 9)} = 7x + 3 \Rightarrow$$ $$(e^y)^{\ln(5x – 9)} = 7x + 3 \Rightarrow$$ $$e^y = (7x + 3)^{\frac{1}{\ln(5x – 9)}} \Rightarrow$$ $$\ln(e^y) = \ln((7x + 3)^{\frac{1}{\ln(5x – 9)}}) \Rightarrow$$ $$y = \frac{\ln(7x + 3)}{\ln(5x – 9)}$$
Now would you want to differentiate that? It’s doable, but the algebra is extremely tricky, and you are likely to trip up somewhere. Implicit differentiation removes much of the algebra by differentiating the equation in a non-explicit form (known as an implicit form).
Implicit differentiation can be performed on an equation in any form besides its explicit forms.
The process of implicit differentiation itself is best illustrated with examples.
Basic Implicit Differentiation
Generally the main objective of an implicit differentiation problem is to find \(\frac{dy}{dx}\) in terms of x and y (although there are other things that can be accomplished with implicit differentiation that will be covered later). This is done by introducing \(\frac{dy}{dx}\) into the equation by differentiating and solving for it.
Example 2: Find \(\frac{dy}{dx}\) where the following equation is true:
$$x + 2y + 7xy + 18 = 0$$
The key to implicit differentiation is to differentiate each term individually as if you were doing explicit differentiation. Since the target is \(\frac{dy}{dx}\), we differentiate with respect to x:
$$1 + \frac{d}{dx}(2y + 7xy) = 0$$
When differentiating a term with \(y\) in it, the \(y\) becomes a \(\frac{dy}{dx}\) since we are differentiating with respect to x. In this particular example, a factoring trick can be used to ensure that only one \(\frac{dy}{dx}\) term appears, making it much easier to solve for once the differentiation is complete:
$$1 + \frac{d}{dx}(y(2 + 7x)) = 0$$
Now use the Product Rule to finish the differentiation (differentiating \(y\) is the Chain Rule in action):
$$1 + \frac{dy}{dx}(2 + 7x) + 7y = 0$$
We can now solve for \(\frac{dy}{dx}\) as if it were any other variable. From here it is a simple algebra problem.
$$\frac{dy}{dx}(2 + 7x) = -7y – 1 \Rightarrow$$ $$\frac{dy}{dx} = \frac{-7y – 1}{7x + 2}$$
Finally note that this relationship does not hold for \(x = -\frac{2}{7}\), because then the denominator of the right side is 0, and the entire right-hand side is undefined.
In general, remember the point that was emphasized in the solution for Example 2:
$$\frac{d}{dx}(y(x)) = \frac{dy}{dx}$$
Here is another example:
Example 3: Find \(\frac{dy}{dx}\) where
$$xy^2 + x^2y – 2x = 12$$
Solution: Again, use the Product Rule and the Chain Rule to differentiate the equation:
$$y^2 + 2xy\frac{dy}{dx} + 2xy + x^2\frac{dy}{dx} – 2 = 0$$
Now we can solve for \(\frac{dy}{dx}\):
$$2xy\frac{dy}{dx} + x^2\frac{dy}{dx} = -y^2 – 2xy + 2 \Rightarrow$$ $$\frac{dy}{dx}(2xy + x^2) = -y^2 – 2xy + 2 \Rightarrow$$ $$\frac{dy}{dx} = \frac{-y^2 – 2xy + 2}{x^2 + 2xy}$$
The Tangent Line Application
The most common application of knowledge of \(\frac{dy}{dx}\) is finding the slope or equation of the tangent line, which adds another level of complexity to the problem.
The slope of the tangent line is equivalent to \(\frac{dy}{dx}\) at that point in the Cartesian plane.
Example 4: Find the slope of the tangent line of
$$x^2 + xy - 144 = y^2$$
when the graph touches the x-axis.
Solution: We want to know what points to find \(\frac{dy}{dx}\) at, so we must solve for \(x\). We want all points where \(y = 0\), so solve for \(x\):
$$x^2 + x(0) – 144 = 0^2 \Rightarrow$$ $$x^2 + 0 -144 = 0 \Rightarrow$$ $$x^2 = 144 \Rightarrow$$ $$x = \pm 12$$
So we want to find \(\frac{dy}{dx}\) at the points \((12, 0)\) and \((-12, 0)\). But in order to do that, we must find the general formula for \(\frac{dy}{dx}\) with implicit differentiation:
$$x^2 + xy – 144 = y^2 \Rightarrow$$ $$2x + y + x\frac{dy}{dx} = 2y\frac{dy}{dx} \Rightarrow$$ $$x\frac{dy}{dx} – 2y\frac{dy}{dx} = -2x – y \Rightarrow$$ $$\frac{dy}{dx}(x – 2y) = -2x – y \Rightarrow$$ $$\frac{dy}{dx} = \frac{-2x – y}{x – 2y} \Rightarrow$$ $$\frac{dy}{dx} = \frac{2x + y}{2y – x}$$
The numerical derivative must be calculated twice, once for each point.
\((12, 0)\):
$$\frac{2(12) + 0}{2(0) – 12} = \frac{24}{-12} = -2$$
\((-12, 0)\):
$$\frac{2(-12) + 0}{2(0) – (-12)} = \frac{-24}{12} = -2$$
So the slope of the tangent line is \(-2\) at both points.
Here is one more [slightly more complicated] example involving tangent lines.
Example 5: Find the equation of the tangent line at \((4, 1)\) in the equation
$$7x + 12y + x^2 = 56$$
Solution: We must begin by finding \(\frac{dy}{dx}\) with implicit differentiation. Differentiate and then solve for \(\frac{dy}{dx}\):
$$7 + 12\frac{dy}{dx} + 2x = 0 \Rightarrow$$ $$12\frac{dy}{dx} = -2x – 7 \Rightarrow$$ $$\frac{dy}{dx} = -\frac{x}{6} - \frac{7}{12}$$
Plug in the coordinates of the given point to find the slope of the line:
$$\frac{dy}{dx} = -\frac{4}{6} - \frac{7}{12} = -\frac{5}{4}$$
The equation of a line in general is
$$y = mx + b$$
We know the slope:
$$y = -\frac{5}{4}x + b$$
Now plug in our coordinate to find the constant \(b\):
$$1 = -\frac{5}{4}(4) + b \Rightarrow$$ $$1 = -5 + b \Rightarrow$$ $$b = 6$$
Therefore the equation is
$$y = -\frac{5}{4}x + 6$$
The Second Derivative
After finding the first derivative, it is possible to use that information to find the second derivative explicitly with some algebraic manipulations. However, algebraically it is more difficult than just finding the first derivative.
Example 6: Find \(\frac{d^2y}{dx^2}\) where
$$x^2 + y^2 – 7xy^2 = 8$$
Solution: As usual, begin by finding \(\frac{dy}{dx}\):
$$2x + 2y\frac{dy}{dx} – 7y^2 – 14xy\frac{dy}{dx} = 0 \Rightarrow$$ $$\frac{dy}{dx}(2y – 14xy) = 7y^2 – 2x \Rightarrow$$ $$\frac{dy}{dx} = \frac{7y^2 – 2x}{2y – 14xy}$$
To find the second derivative, we must differentiate the first derivative (which can be fairly messy).
$$\frac{d^2y}{dx^2} = \frac{(2y – 14xy)\frac{d}{dx}(7y^2 – 2x) – (7y^2 – 2x)\frac{d}{dx}(2y – 14xy)}{(2y – 14xy)^2} \Rightarrow$$ $$\frac{d^2y}{dx^2} = \frac{(2y – 14xy)(14y\frac{dy}{dx} – 2) + (2x – 7y^2)(2\frac{dy}{dx} – 14y – 14x\frac{dy}{dx})}{4y^2 – 56xy^2 + 196x^2y^2} \Rightarrow$$ $$\frac{d^2y}{dx^2} =$$ $$\frac{28y^2\frac{dy}{dx} – 4y – 196xy^2\frac{dy}{dx} + 28xy + 4x\frac{dy}{dx} – 28xy – 28x^2\frac{dy}{dx} – 14y^2\frac{dy}{dx} + 98y^3 + 98xy^2\frac{dy}{dx}}{4y^2 – 56xy^2 + 196x^2y^2} \Rightarrow$$ $$\frac{d^2y}{dx^2} = \frac{14y^2\frac{dy}{dx} – 98xy^2\frac{dy}{dx} – 4y + 98y^3 + 4x\frac{dy}{dx} – 28x^2\frac{dy}{dx}}{(2y – 14xy)^2}$$
We pull out a common factor of \(\frac{dy}{dx}\) wherever it is present:
$$\frac{d^2y}{dx^2} = \frac{\frac{dy}{dx}(14y^2 – 98xy^2 – 28x^2 + 4x) – 4y + 98y^3}{(2y – 14xy)^2}$$
Now divide out a common factor of \(2\):
$$\frac{d^2y}{dx^2} = \frac{\frac{dy}{dx}(7y^2 – 49xy^2 – 14x^2 + 2x) – 2y + 49y^3}{(y – 7xy)(2y – 14xy)}$$
We can substitute \(\frac{dy}{dx}\) into this expression:
$$\frac{d^2y}{dx^2} = \frac{\frac{7y^2 – 2x}{2y – 14xy}(7y^2 – 49xy^2 – 14x^2 + 2x) – 2y + 49y^3}{(y – 7xy)(2y – 14xy)}$$
Rewrite this to eliminate the complex fraction:
$$\frac{d^2y}{dx^2} = \frac{(7y^2 – 2x)(7y^2 – 49xy^2 – 14x^2 + 2x) – (2y – 49y^3)(2y – 14xy)}{(y – 7xy)(2y – 14xy)^2}$$
Since this is only in terms of \(x\) and \(y\), there is no need to simplify this any further.
A Graphical Perspective of Implicit Differentiation
The concept of implicit differentiation can be made more realistic by using graphs to locate points where the derivative is zero or undefined, and use it to check your algebra.
Example 7: We will use the graph of the equation in Example 1:
$$x + 2y + 7xy + 18 = 0$$
As a check, notice that there is a vertical asymptote at \(x = -\frac{2}{7}\), at which the derivative is undefined. It is common sense that when a function is undefined at a point, its derivative will be as well.
There are many instances where implicit differentiation is helpful, primarily where explicit differentiation is too complicated algebraically to be feasibly executed. But since it is still differentiation, all the applications remain the same.