In the previous sections we learned to find the derivative, \(\lz{y}{x}\text{,}\) or \(y'\text{,}\) when \(y\) is given explicitly as a function of \(x\text{.}\) That is, if we know \(y=f(x)\) for some function \(f\text{,}\) we can find \(y'\text{.}\) For example, given \(y=3x^2-7\text{,}\) we can easily find \(y'=6x\text{.}\) (Here we explicitly state how \(y\) depends on \(x\text{.}\) Knowing \(x\text{,}\) we can directly find \(y\text{.}\))
Sometimes the relationship between \(y\) and \(x\) is not explicit; rather, it is implicit. For instance, we might know that \(x^2-y=4\text{.}\) This equality defines a relationship between \(x\) and \(y\text{;}\) if we know \(x\text{,}\) we could figure out \(y\text{.}\) Can we still find \(y'\text{?}\) In this case, sure; we solve for \(y\) to get \(y=x^2-4\) (hence we now know \(y\) explicitly) and then differentiate to get \(y'=2x\text{.}\)
Sometimes the implicit relationship between \(x\) and \(y\) is complicated. Suppose we are given \(\sin(y)+y^3=6-x^3\text{.}\) A graph of this implicit relationship is given in Figure 2.6.2. In this case there is absolutely no way to solve for \(y\) in terms of elementary functions. The surprising thing is, however, that we can still find \(y'\) via a process known as implicit differentiation.
The curve begins in the second quadrant. From the left, the curve decreases as \(x\) increases. The curve slowly flattens out, almost becoming horizontal as the curve crosses the \(y\)-axis near the point \((0,1.8)\text{.}\) When \(x\) 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 \((0,1.8)\text{,}\) at which it also crosses into the fourth quadrant. When \(x\) 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).
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
Implicit functions are generally harder to deal with than explicit functions. With an explicit function, given an \(x\) value, we have an explicit formula for computing the corresponding \(y\) value. With an implicit function, one often has to find \(x\) and \(y\) 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 \(\left(\sqrt[3]{6},0\right)\) lies on the graph of the implicit function \(\sin(y) + y^3=6-x^3\text{.}\) Plugging in \(0\) for \(y\text{,}\) we see the left hand side is \(0\text{.}\) Setting \(x=\sqrt[3]6\text{,}\) we see the right hand side is also \(0\text{;}\) the equation is satisfied. The following example finds the equation of the tangent line to this function at this point.
Example2.6.4.Using implicit differentiation to find a tangent line.
Find the equation of the line tangent to the curve of the implicitly defined function \(\sin(y) + y^3=6-x^3\) at the point \(\left(\sqrt[3]6,0\right)\text{.}\)
We find the slope of the tangent line at the point \(\left(\sqrt[3]6,0\right)\) by substituting \(\sqrt[3]6\) for \(x\) and \(0\) for \(y\text{.}\) Thus at the point \(\left(\sqrt[3]6,0\right)\text{,}\) we have the slope as
Take the derivative of each term in the equation. Treat the \(x\) terms like normal. When taking the derivatives of \(y\) terms, the usual rules apply except that, because of the Theorem 2.5.4, we need to multiply each term by \(y'\text{.}\)
(Practical Note: when working by hand, it may be beneficial to use the symbol \(\frac{dy}{dx}\) instead of \(y'\text{,}\) as the latter can be easily confused for \(y\) or \(y^1\text{.}\))
The second term, \(x^2y^4\text{,}\) is a little tricky. It requires the Product Rule as it is the product of two functions of \(x\text{:}\)\(x^2\) and \(y^4\text{.}\) Its derivative is \(x^2(4y^3y') + 2xy^4\text{.}\) The first part of this expression requires a \(y'\) because we are taking the derivative of a \(y\) term. The second part does not require it because we are taking the derivative of \(x^2\text{.}\)
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 \((0,1)\) lies on the graph of this function. At this point, \(y' = 2/3\text{.}\) So the equation of the tangent line is \(y = 2/3(x-0)+1\text{.}\) The function and its tangent line are graphed in Figure 2.6.7.
Two curves are drawn in the \(xy\)-plane. The left curve stretches upwards from the left side of the \(y\) axis, curving slightly to the left. As \(y\) approaches -2, the curve begins to widen to the left, creating a bump in the curve. As the curve crosses the \(x\) axis, the curve moves towards the right, no longer increasing and becoming more horizontal as \(x\) increases. At the point \((0,1)\text{,}\) 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 \(y\)-axis, as a line stretching upwards from the bottom of the \(y\)-axis. As \(x\) approaches 1, the curve also begins to become horizontal as \(x\) increases. The entire second curve lies in the fourth quadrant.
Figure2.6.7.A graph of the implicitly defined function \(y^3+x^2y^4=1+2x\) along with its tangent line at the point \((0,1)\)
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 \(y\) as a function of \(x\text{.}\) But when we indicate we are interested in the derivative at \((0,1)\text{,}\) we are indicating that we want the function defined by the small portion of the curve that passes through \((0,1)\text{,}\) 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.
We now have to be careful to properly solve for \(y'\text{,}\) particularly because of the product on the left. It is best to multiply out the product. Doing this, we get
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 \(y\)-axis at \(y = -1\) into the fourth quadrant. The curve continues to increase as such until it reaches a point close to \((\frac{1}{2},-\frac{3}{4}\text{.}\) 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 \(y\)-axis at \(y = 1\) into the first quadrant. From there, the curve continues towards the right while slightly increasing. The curves rises sharply at \(x = 1.5\text{,}\) before decreasing again.
It is easy to verify that the points \((0,0)\text{,}\)\((0,1)\) and \((0,-1)\) all lie on the graph. We can find the slopes of the tangent lines at each of these points using our formula for \(y'\text{.}\)
The graph in Figure 2.6.9, with tagent lines drawn at \((0,-1)\text{,}\)\((0,0)\text{,}\) and \((0,1)\text{.}\) The tangent line at \((0,-1)\) has a positive slope less than 1. The tangent line at \((0,0)\) has a negative slope, close to -1. The tangent line at \((0,1)\) has a positive slope, less than 1.
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 \((x,y)\) on the circle will be \(y/x\text{.}\) We have found that the slope of the tangent line to the circle at that point is the opposite reciprocal of \(y/x\text{,}\) namely, \(-x/y\text{.}\) Hence these two lines are always perpendicular.
A graph of the circle and its tangent line at \(\left(1/2,\sqrt{3}/2\right)\) 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 \((\frac{1}{2},\frac{\sqrt{3}}{2})\text{.}\) At that point a tangent line is drawn with a slight negative slope.
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 \(y = x^{m/n}\text{,}\) where \(m\) and \(n\) are integers with no common factors (so \(m=2\) and \(n=5\) is fine, but \(m=2\) and \(n=4\) is not). We can rewrite this explicit function implicitly as \(y^n = x^m\text{.}\) 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.
Let \(f(x) = x^n\text{,}\) where \(n\neq 0\) is a real number. Then \(f\) is differentiable on its domain, except possibly at \(x=0\text{,}\) and \(\fp(x) = n\cdot x^{n-1}\text{.}\)
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.
Plugging in \(x=8\) and \(y=8\text{,}\) we get a slope of \(-1\text{.}\) The astroid, with its tangent line at \((8,8)\text{,}\) is shown in Figure 2.6.16.
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 \(\lz{y}{x}\text{,}\) then take its derivative with respect to \(x\text{.}\) In practice, it is not hard, but it often requires a bit of algebra. We demonstrate this in an example.
While this is not a particularly simple expression, it is usable. We can see that \(y'' \gt 0\) when \(y\lt 0\) and \(y''\lt 0\) when \(y \gt 0\text{.}\) 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 \(x^2+y^2=1\text{,}\) then we know that \(x^2=1-y^2\text{.}\) So we can replace the \(x^2\) in the expression for \(y''\) to get
Consider the function \(y=x^x\text{;}\) it is graphed in Figure 2.6.18. It is well-defined for \(x \gt 0\) 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 \((0,1)\) there is a discontinuity. The curve begins decreasing, reaching a minimum at around \((0.3,0.7)\text{.}\) After that point, the curve increases exponentially.
The function is not a power function: it has a “power” of \(x\text{,}\) not a constant. It is not an exponential function either: it has a “base” of \(x\text{,}\) 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 \(y=f(x)\text{,}\) then use implicit differentiation to find \(y'\text{.}\) We demonstrate this in the following example.
To “test” our answer, let’s use it to find the equation of the tangent line at \(x=1.5\text{.}\) The point on the graph our tangent line must pass through is \(\left(1.5, 1.5^{1.5}\right) \approx (1.5, 1.837)\text{.}\) Using the equation for \(y'\text{,}\) we find the slope as
Thus the equation of the tangent line is (approximately) \(y \approx 2.582(x-1.5)+1.837\text{.}\)Figure 2.6.20 graphs \(y=x^x\) along with this tangent line.
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 \(y=\sin^{-1}(x)\text{.}\)
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.
