In the previous sections we learned to find the derivative, , or , when is given explicitly as a function of . That is, if we know for some function , we can find . For example, given , we can easily find . (Here we explicitly state how depends on . Knowing , we can directly find .)
Sometimes the relationship between and is not explicit; rather, it is implicit. For instance, we might know that . This equality defines a relationship between and ; if we know , we could figure out . Can we still find ? In this case, sure; we solve for to get (hence we now know explicitly) and then differentiate to get .
Sometimes the implicit relationship between and is complicated. Suppose we are given . A graph of this implicit relationship is given in Figure 2.6.2. In this case there is absolutely no way to solve for in terms of elementary functions. The surprising thing is, however, that we can still find via a process known as implicit differentiation.
The curve begins in the second quadrant. From the left, the curve decreases as increases. The curve slowly flattens out, almost becoming horizontal as the curve crosses the -axis near the point . When comes close to 0.75, the curve begins decreasing in the shape of a gentle corner. The curve continues decreasing, becoming steepest around the point , at which it also crosses into the fourth quadrant. When is close to 2, the curve begins to decrease more gently, at around the same rate as the beginning of the curve.
Subsection2.6.1The method of implicit differentiation
Implicit differentiation is a technique based on the The Chain Rule that is used to find a derivative when the relationship between the variables is given implicitly rather than explicitly (solved for one variable in terms of the other).
We start by taking the derivative of both sides (thus maintaining the equality.) We have:
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The right hand side is easy; it returns .
The left hand side requires more consideration. We take the derivative term-by-term. Using the technique derived from Equation (2.6.1) above, we can see that
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We apply the same process to the term.
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Putting this together with the right hand side, we have
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Now solve for . It’s important to treat as an algebraically independent variable from and .
This equation for probably seems unusual for it contains both and terms. How is it to be used? We’ll address that next.
Implicit functions are generally harder to deal with than explicit functions. With an explicit function, given an value, we have an explicit formula for computing the corresponding value. With an implicit function, one often has to find and values at the same time that satisfy the equation. It is much easier to demonstrate that a given point satisfies the equation than to actually find such a point.
For instance, we can affirm easily that the point lies on the graph of the implicit function . Plugging in for , we see the left hand side is . Setting , we see the right hand side is also ; the equation is satisfied. The following example finds the equation of the tangent line to this function at this point.
Take the derivative of each term in the equation. Treat the terms like normal. When taking the derivatives of terms, the usual rules apply except that, because of the Theorem 2.5.4, we need to multiply each term by .
Get all the terms on one side of the equal sign and put the remaining terms on the other side.
We will take the implicit derivatives term by term. The derivative of is .
The second term, , is a little tricky. It requires the Product Rule as it is the product of two functions of : and . Its derivative is . The first part of this expression requires a because we are taking the derivative of a term. The second part does not require it because we are taking the derivative of .
The derivative of the right hand side is easily found to be . In all, we get:
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Move terms around so that the left side consists only of the terms and the right side consists of all the other terms:
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Factor out from the left side and solve to get
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To confirm the validity of our work, let’s find the equation of a tangent line to this function at a point. It is easy to confirm that the point lies on the graph of this function. At this point, . So the equation of the tangent line is . The function and its tangent line are graphed in Figure 2.6.7.
Two curves are drawn in the -plane. The left curve stretches upwards from the left side of the axis, curving slightly to the left. As approaches -2, the curve begins to widen to the left, creating a bump in the curve. As the curve crosses the axis, the curve moves towards the right, no longer increasing and becoming more horizontal as increases. At the point , a tangent line is drawn, with a moderate positive slope. This point corresponds to the corner at which the curve begins to become horizontal. At this point, the curve passes the vertical line test, but does not at most other points on the graph. The second curve begins to the right of the -axis, as a line stretching upwards from the bottom of the -axis. As approaches 1, the curve also begins to become horizontal as increases. The entire second curve lies in the fourth quadrant.
Figure2.6.7.A graph of the implicitly defined function along with its tangent line at the point
Notice how our curve looks much different than for functions we have seen. For one, it fails the vertical line test, and so the complete curve is not truly representing as a function of . But when we indicate we are interested in the derivative at , we are indicating that we want the function defined by the small portion of the curve that passes through , and that small portion does pass the vertical line test. Such functions are important in many areas of mathematics, so developing tools to deal with them is also important.
Differentiating term by term, we find the most difficulty in the first term. It requires both the The Chain Rule and Product Rule.
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We leave the derivatives of the other terms to the reader. After taking the derivatives of both sides, we have
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We now have to be careful to properly solve for , particularly because of the product on the left. It is best to multiply out the product. Doing this, we get
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From here we can safely move around terms to get the following:
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Then we can solve for to get
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A graph of this implicit function is given in Figure 2.6.9.
The curve begins in the third quadrant. From there, the curve bends slightly back and increases, crossing above itself. The curve extends to the right, increasing almost linearly as it crosses the -axis at into the fourth quadrant. The curve continues to increase as such until it reaches a point close to . The curve then bends back, increasing towards the top left linearly. It then crosses the origin and passes into the second quadrant. The curve quickly bends towards the right, crossing the -axis at into the first quadrant. From there, the curve continues towards the right while slightly increasing. The curves rises sharply at , before decreasing again.
Figure2.6.9.A graph of the implicitly defined curve
It is easy to verify that the points , and all lie on the graph. We can find the slopes of the tangent lines at each of these points using our formula for .
At , the slope is .
At , the slope is .
At , the slope is also .
The tangent lines have been added to the graph of the function in Figure 2.6.10.
The graph in Figure 2.6.9, with tagent lines drawn at ,, and . The tangent line at has a positive slope less than 1. The tangent line at has a negative slope, close to -1. The tangent line at has a positive slope, less than 1.
Figure2.6.10.A graph of the implicitly defined curve and certain tangent lines
Quite a few “famous” curves have equations that are given implicitly. We can use implicit differentiation to find the slope at various points on those curves. We investigate two such curves in the next examples.
This is a clever formula. Recall that the slope of the line through the origin and the point on the circle will be . We have found that the slope of the tangent line to the circle at that point is the opposite reciprocal of , namely, . Hence these two lines are always perpendicular.
At the point , we have the tangent line’s slope as
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A graph of the circle and its tangent line at is given in Figure 2.6.12, along with a thin dashed line from the origin that is perpendicular to the tangent line. (It turns out that all normal lines to a circle pass through the center of the circle.)
A circle of radius 1 centered at the origin. A dashed line extends from the origin to a tangent line at the point . At that point a tangent line is drawn with a slight negative slope.
Figure2.6.12.The unit circle with its tangent line at
This section has shown how to find the derivatives of implicitly defined functions, whose graphs include a wide variety of interesting and unusual shapes. Implicit differentiation can also be used to further our understanding of “regular” differentiation.
We allude to a possible solution, as we can write the square root function as a power function with a rational (or, fractional) power. We are then tempted to apply the Power Rule with Integer Exponents and obtain
Let , where and are integers with no common factors (so and is fine, but and is not). We can rewrite this explicit function implicitly as . Now apply implicit differentiation.
The above derivation is the key to the proof extending the Power Rule with Integer Exponents to rational powers. Using limits, we can extend this once more to include all powers, including irrational (even transcendental!) powers, giving the following theorem.
This is a particularly interesting curve called an astroid. It is the shape traced out by a point on the edge of a circle that is rolling around inside of a larger circle, as shown in Figure 2.6.15.
A dashed circle of radius 20 entirely contains the curve. In each quadrant curves connect the points on the x and y axis which also lie on the circle. This gives the overall curve the appearence of a diamond with sides curved towards the inside. In the third quadrant a smaller circle is drawn which touches both the outer circle and the curve. The point on the circle touching the curve is highlighted blue.
Figure2.6.15.An astroid, traced out by a point on the smaller circle as it rolls inside the larger circle
To find the slope of the astroid at the point , we take the derivative implicitly.
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Plugging in and , we get a slope of . The astroid, with its tangent line at , is shown in Figure 2.6.16.
The curve sketched in Figure 2.6.15 with a tangent line at . It has a slope of -1.
Subsection2.6.2Implicit Differentiation and the Second Derivative
We can use implicit differentiation to find higher order derivatives. In theory, this is simple: first find , then take its derivative with respect to . In practice, it is not hard, but it often requires a bit of algebra. We demonstrate this in an example.
We found that in Example 2.6.11. To find , we apply implicit differentiation to .
(Now use the Quotient Rule.)replace with :.
While this is not a particularly simple expression, it is usable. We can see that when and when . In Section 3.4, we will see how this relates to the shape of the graph.
Also, if we remember that we are only considering points on the curve , then we know that . So we can replace the in the expression for to get
which is a simpler expression. Recognizing when simplifications like this are possible is not always easy.
Consider the function ; it is graphed in Figure 2.6.18. It is well-defined for and we might be interested in finding equations of lines tangent and normal to its graph. How do we take its derivative?
The curve is entirely contained within the first quadrant. At the point there is a discontinuity. The curve begins decreasing, reaching a minimum at around . After that point, the curve increases exponentially.
The function is not a power function: it has a “power” of , not a constant. It is not an exponential function either: it has a “base” of , not a constant.
A differentiation technique known as logarithmic differentiation becomes useful here. The basic principle is this: take the natural log of both sides of an equation , then use implicit differentiation to find . We demonstrate this in the following example.
As suggested above, we start by taking the natural log of both sides then applying implicit differentiation.
(apply logarithm rule)(now use implicit differentiation)(substitute ).
To “test” our answer, let’s use it to find the equation of the tangent line at . The point on the graph our tangent line must pass through is . Using the equation for , we find the slope as
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Thus the equation of the tangent line is (approximately) .Figure 2.6.20 graphs along with this tangent line.
The graph shown in Figure 2.6.18, with a tangent line at . It has a slope of around 2.6.
We would not have been able to compute the derivative of the function in Example 2.6.19 without logarithmic differentiation. But the method is also useful in cases where the product and quotient rules could be used, but logarithmic differentiation is simpler. The video in Figure 2.6.21 provides such an example.
Implicit differentiation proves to be useful as it allows us to find the instantaneous rates of change of a variety of functions. In particular, it extended the Power Rule for Differentiation to rational exponents, which we then extended to all real numbers. In Section 2.7, implicit differentiation will be used to find the derivatives of inverse functions, such as .
Find the equation of the tangent line to the graph of the implicitly defined function at the indicated points. As a visual aid, the function is graphed.
