2.7: Implicit Differentiation and the Derivatives of the Inverse Trigonometric and Hyperbolic Functions

Learning Objectives Use implicit differentiation to find the derivative given an implicitly defined relation between two variables. Use implicit differentiation to determine the equation of a tangent line to an implicitly-defined curve. Calculate the derivative of an inverse trigonometric function. Recognize the derivatives of the inverse hyperbolic functions.

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We have already studied how to find equations of tangent lines to functions and the rate of change of a function at a specific point. In all these cases, we had the explicit equation for the function and differentiated these functions explicitly. Suppose instead that we want to determine the equation of a tangent line to an arbitrary curve or the rate of change of an arbitrary curve at a point. In this section, we solve these problems by finding the derivatives of functions that define \(y\) implicitly in terms of \(x\).

Implicit Differentiation

In most discussions of math, if the dependent variable \(y\) is a function of the independent variable \(x\), we express \(y\) in terms of \(x\). If this is the case, we say that \(y\) is an explicit function of \(x\). For example, when we write the equation \(y=x^2+1\), we are defining \(y\) explicitly in terms of \(x\). On the other hand, if the relationship between the function \(y\) and the variable \(x\) is expressed by an equation where \(y\) is not expressed entirely in terms of \(x\), we say that the equation defines \(y\) implicitly in terms of \(x\). For example, the equation \(y−x^2=1\) defines the function \(y=x^2+1\) implicitly.

Implicit differentiation allows us to find slopes of tangents to curves that are not functions (they fail the Vertical Line Test). We are using the idea that portions of \(y\) are functions that satisfy the given equation but that \(y\) is not a function of \(x\).

Generally, an equation defines a function implicitly if the function satisfies that equation. An equation may define many different functions implicitly. For example, the functions\[y=\sqrt{25−x^2}\nonumber \]and\[y=\begin{cases}\sqrt{25−x^2}, & \text{if }−5 \leq x<0\\ −\sqrt{25−x^2}, & \text{if }0 \leq x \leq 5\end{cases}\nonumber \]which are illustrated in Figure \(\PageIndex{1}\), are just two of the many functions defined implicitly by the equation \(x^2+y^2=25\).

Figure \(\PageIndex{1}\):The equation \(x^2+y^2=25\) defines many functions implicitly.

If we want to find the slope of the line tangent to the graph of \(x^2+y^2=25\) at the point \((3,4)\), we could evaluate the derivative of the function \(y=\sqrt{25−x^2}\) at \(x=3\). On the other hand, if we want the slope of the tangent line at the point \((3,−4)\), we could use the derivative of \(y=−\sqrt{25−x^2}\). However, it is not always easy to solve for a function defined implicitly by an equation. Fortunately, the technique of implicit differentiation allows us to find the derivative of an implicitly defined function without ever solving for the function explicitly.

A simplified explanation of implicit differentiation is that you take the derivatives of both sides of a given equation (whether explicitly solved for \(y\) or not) with respect to the independent variable and perform the Chain Rule whether or not it is necessary. For example, suppose \(y = \sinh{(x)} - 2^x\). Then\[ \begin{array}{rcl}
\dfrac{d}{dx} \left( y \right) = \dfrac{d}{dx} \left( \sinh{(x)} - 2^x \right) & \implies & \dfrac{dy}{dx} = \cosh{(x)} \dfrac{dx}{dx} - 2^x \ln{(2)} \dfrac{dx}{dx} \\[16pt]
& \implies & \dfrac{dy}{dx} = \cosh{(x)} \cdot 1 - 2^x \ln{(x)} \cdot 1 \\[16pt]
& \implies & \dfrac{dy}{dx} = \cosh{(x)} - 2^x \ln{(x)} \\[16pt]
\end{array} \nonumber \]While the results are no different than how we have performed differentiation so far, the process is important to understand. Suppose, instead, that we had the equation \(\cos{(x^2+x)} = \tanh{(x)} + 3y\) and were asked to find \(y^{\prime}\). We could start by solving for \(y\), or we could take the derivative of both sides immediately as follows:\[ \begin{array}{rcl}
\dfrac{d}{dx} \left( \cos{(x^2+x)} \right) = \dfrac{d}{dx} \left( \tanh{(x)} + 3y \right) & \implies & -\sin{(x^2 +x)} \cdot \dfrac{d}{dx} \left( x^2 + x\right) = \text{sech}^2(x) \dfrac{dx}{dx} + 3 \dfrac{dy}{dx} \\[16pt]
& \implies & -\sin{(x^2 +x)} \cdot \left( 2x + 1\right) \cdot \dfrac{dx}{dx} = \text{sech}^2(x) \cdot 1 + 3 \dfrac{dy}{dx} \\[16pt]
& \implies & -\sin{(x^2 +x)} \cdot \left( 2x + 1\right) \cdot 1 = \text{sech}^2(x) \cdot 1 + 3 \dfrac{dy}{dx} \\[16pt]
& \implies & \dfrac{-\sin{(x^2 +x)} \cdot \left( 2x + 1\right) - \text{sech}^2(x)}{3} = \dfrac{dy}{dx} \\[16pt]
\end{array} \nonumber \]However, the real power of implicit differentiation comes into play when given the equation for an implicitly defined curve. For example, if \(e^{xy} - \sqrt{x + y} = \cot{(y)}\), then the derivative of \(y\) with respect to \(x\) can be found by taking the derivative implicitly.\[ \begin{array}{rcl}
\dfrac{d}{dx} \left( e^{xy} - \sqrt{x + y} \right) = \dfrac{d}{dx} \left( \cot{(y)} \right) & \implies & \dfrac{d}{dx} e^{xy} - \dfrac{d}{dx}\sqrt{x + y} = -\csc^2{(y)} \cdot \dfrac{dy}{dx} \\[16pt]
& \implies & e^{xy} \cdot \dfrac{d}{dx} \left( xy \right) - \dfrac{1}{2 \sqrt{x + y}} \cdot \dfrac{d}{dx} \left( x + y \right) = -\csc^2{(y)} \cdot \dfrac{dy}{dx} \\[16pt]
& \implies & e^{xy} \cdot \left(\dfrac{dx}{dx} \cdot y + x \cdot \dfrac{dy}{dx} \right) - \dfrac{1}{2 \sqrt{x + y}} \cdot \left( \dfrac{dx}{dx} + \dfrac{dy}{dx} \right) = -\csc^2{(y)} \cdot \dfrac{dy}{dx} \\[16pt]
& \implies & e^{xy} \cdot \left(y + x \cdot \dfrac{dy}{dx} \right) - \dfrac{1}{2 \sqrt{x + y}} \cdot \left( 1 + \dfrac{dy}{dx} \right) = -\csc^2{(y)} \cdot \dfrac{dy}{dx} \\[16pt]
\end{array} \nonumber \]At this point, we would feel compelled to solve for \(\frac{dy}{dx}\); however, let's hold off for now. It's enough at this point to understand how implicit differentiation works.

Example \(\PageIndex{1}\): Using Implicit Differentiation

Assuming that \(y\) is defined implicitly by the equation \(x^2+y^2=25\), find \(\frac{dy}{dx}\).

Solution

\[ \begin{array}{rrclcl}
& \dfrac{d}{dx}(x^2+y^2) & = & \dfrac{d}{dx}(25) & \quad & \left( \text{differentiating implicitly} \right) \\[16pt]
\implies & \dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(y^2) & = & 0 & \quad & \left( \text{Sum and Constant derivative rules} \right) \\[16pt]
\implies & 2x+2y\dfrac{dy}{dx} & = & 0 & \quad & \left( \dfrac{d}{dx}(x^2)=2x \, \dfrac{dx}{dx} = 2x \text{ and }\dfrac{d}{dx}(y^2)=2y \, \dfrac{dy}{dx} \right) \\[16pt]
\implies & 2y\dfrac{dy}{dx} & = & −2x & \quad & \left( \text{grouping like terms} \right) \\[16pt]
\implies & \dfrac{dy}{dx} & = & −\dfrac{x}{y} & \quad & \left( \text{dividing both sides by }2y \right) \\[16pt]
\end{array} \nonumber \]

Analysis

Note that the resulting expression for \(\frac{dy}{dx}\) is in terms of both the independent variable \(x\) and the dependent variable \(y\). Although in some cases it may be possible to express \(\frac{dy}{dx}\) in terms of \(x\) only, it is generally not possible to do so.

Example \(\PageIndex{2}\): Using Implicit Differentiation and the Product Rule

Assuming that \(y\) is defined implicitly by the equation \(x^3\sin y+y=4x+3\), find \(\frac{dy}{dx}\).

Solution

\[ \begin{array}{rrclcl}
& \dfrac{d}{dx}(x^3\sin y+y) & = & \dfrac{d}{dx}(4x+3) & \quad & \left( \text{differentiating implicitly} \right) \\[16pt]
\implies & \dfrac{d}{dx}(x^3\sin y)+\dfrac{d}{dx}(y) & = & 4 & & \left( \text{Sum Rule on the left, and }\dfrac{d}{dx}(4x+3)=4 \text{ on the right} \right) \\[16pt]
\implies & \left(\dfrac{d}{dx}(x^3) \cdot \sin y+\dfrac{d}{dx}(\sin y) \cdot x^3\right)+\dfrac{dy}{dx} & = & 4 & \quad & \left( \text{Product Rule} \right) \\[16pt]
\implies & 3x^2\sin y+(\cos y\dfrac{dy}{dx}) \cdot x^3+\dfrac{dy}{dx} & = & 4 & \quad & \left( \dfrac{d}{dx}(x^3)=3x^2, \text{ and Chain Rule to obtain }\dfrac{d}{dx}(\sin y)=\cos y\dfrac{dy}{dx} \right) \\[16pt]
\implies & x^3\cos y\dfrac{dy}{dx}+\dfrac{dy}{dx} & = & 4−3x^2\sin y & \quad & \left( \text{grouping like terms} \right) \\[16pt]
\implies & \dfrac{dy}{dx}(x^3\cos y+1) & = & 4−3x^2\sin y & \quad & \left( \text{factoring} \right) \\[16pt]
\implies & \dfrac{dy}{dx} & = & \dfrac{4−3x^2\sin y}{x^3\cos y+1} & \quad & \left( \text{solving for }\dfrac{dy}{dx} \right) \\[16pt]
\end{array} \nonumber \]

Example \(\PageIndex{3}\): Using Implicit Differentiation to Find a Second Derivative

Find \(\frac{d^2y}{dx^2}\) if \(x^2+y^2=25\).

Solution

In Example \(\PageIndex{1}\), we showed that \(\frac{dy}{dx}=−\frac{x}{y}\). We can take the derivative of both sides of this equation to find \(\frac{d^2y}{dx^2}\).\[\begin{array}{rclcl}
\dfrac{d^2y}{dx^2} & = & \dfrac{d}{dy}\left(−\dfrac{x}{y}\right) & \quad & \left( \text{differentiating implicitly} \right)\\[16pt]
& = & −\dfrac{\left(1 \cdot y−x\dfrac{dy}{dx}\right)}{y^2} & \quad & \left( \text{Quotient Rule} \right) \\[16pt]
& = & \dfrac{−y+x\dfrac{dy}{dx}}{y^2} & & \\[16pt]
& = & \dfrac{−y+x\left(−\dfrac{x}{y}\right)}{y^2} & \quad & \left( \text{substituting }\dfrac{dy}{dx}=−\dfrac{x}{y}. \right) \\[16pt]
& = & \dfrac{−y^2−x^2}{y^3} & & \\[16pt]
\end{array}\nonumber\]At this point we have found an expression for \(\frac{d^2y}{dx^2}\). If we choose, we can simplify the expression further by recalling that \(x^2+y^2=25\) and making this substitution in the numerator to obtain \(\frac{d^2y}{dx^2}=−\frac{25}{y^3}\).

Checkpoint \(\PageIndex{3}\)

Find \(\frac{dy}{dx}\) for \(y\) defined implicitly by the equation \(4x^5+\tan y=y^2+5x\).

Answer

\[\dfrac{dy}{dx}=\dfrac{5−20x^4}{\sec^2y−2y} \nonumber \]

Finding Tangent Lines Implicitly

Now that we have seen the technique of implicit differentiation, we can apply it to the problem of finding equations of tangent lines to curves described by equations.

Great Advice to Save Time and Prevent Mistakes

When asked to find the value of the derivative or the equation of the tangent line for an implicitly-defined curve at a given point, it's best to not solve for \( \frac{dy}{dx} \) immediately after implicitly differentiating. Instead, once you have implicitly differentiated the relation, substitute the given values of \( x \) and \( y \) into the resulting equation. It is at that point that you should solve for \( \frac{dy}{dx} \). Examples \( \PageIndex{4} \) and \( \PageIndex{5} \) illustrate this process.

Example \(\PageIndex{4}\): Finding the Equation of a Tangent Line

Find the equation of the line tangent to the curve \(\cosh{(xy)} + xy = \frac{5}{4} + \ln{(2)}\) at the point \((1,\ln{(2)})\).

Solution

\[ \begin{array}{rrclcl}
& \dfrac{d}{dx}\left( \cosh{(xy)} + xy \right) & = & \dfrac{d}{dx}\left( \dfrac{5}{4} + \ln{(2)} \right) & & \\[16pt]
\implies & \dfrac{d}{dx}\cosh{(xy)} + \dfrac{d}{dx} \left( xy \right) & = & 0 & \quad & \left( \text{Sum and Constant Rules} \right) \\[16pt]
\implies & \sinh{(xy)} \cdot \dfrac{d}{dx}\left(xy\right) + \dfrac{d}{dx} \left( xy \right) & = & 0 & \quad & \left( \text{Chain Rule} \right) \\[16pt]
\implies & \sinh{(xy)} \cdot \left( \dfrac{d}{dx}x \cdot y + x \cdot \dfrac{d}{dx} y \right) + \left( \dfrac{d}{dx}x \cdot y + x \cdot \dfrac{d}{dx} y \right) & = & 0 & \quad & \left( \text{Product Rule} \right) \\[16pt]
\implies & \sinh{(xy)} \cdot \left( y + x \cdot \dfrac{dy}{dx} \right) + \left( y + x \cdot \dfrac{dy}{dx} \right) & = & 0 & \quad & \left( \text{Power Rule} \right) \\[16pt]
\end{array} \nonumber \]At this point, it is important to remember the task at hand. We are asked to find the equation of the tangent line to the curve \(\cosh{(xy)} + xy = \frac{5}{4} + \ln{(2)}\) at the point \((1,\ln{(2)})\). There is no need to explicitly solve for \( \frac{dy}{dx} \) at this moment. Instead, let's substitute \( 1 \) for \( x \) and \( \ln{(2)} \) for \( y \).\[ \begin{array}{rrclcl}
& \sinh{(\ln{(2)})} \cdot \left( \ln{(2)} + \dfrac{dy}{dx} \right) + \left( \ln{(2)} + \dfrac{dy}{dx} \right) & = & 0 & \quad & \left( \text{substituting} \right) \\[16pt]
\implies & \left( \sinh{(\ln{(2)})} + 1 \right) \cdot \left( \ln{(2)} + \dfrac{dy}{dx} \right) & = & 0 & & \\[16pt]
\implies & \left( \ln{(2)} + \dfrac{dy}{dx} \right) & = & 0 & \quad & \left( \text{dividing both sides by a nonzero constant}^* \right) \\[16pt]
\implies & \dfrac{dy}{dx}\Big|_{(1, \ln{(2)})} & = & -\ln{(2)} & & \\[16pt]
\end{array} \nonumber \]Now that we have the slope of the tangent line at the point \( (1, \ln{(2)} \), we use the point-slope form of a line to get the equation of the tangent line.\[ y - \ln{(2)} = -\ln{(2)}(x - 1) \implies y = -\ln{(2)}x + 2 \ln{(2)}. \nonumber \]Figure \( \PageIndex{2} \) shows the curve and the tangent line.

Figure \( \PageIndex{2} \)

^* I leave it to the reader to check that \( \sinh{(\ln{(2)})} + 1 \neq 0 \).

Example \(\PageIndex{5}\): Finding the Equation of the Tangent Line to a Curve

Find the equation of the line tangent to the graph of \(y^3+x^3−3xy=0\) at the point \(\left(\frac{3}{2},\frac{3}{2}\right)\) (Figure \(\PageIndex{3}\)). This curve is known as the folium (or leaf) of Descartes.

Figure \(\PageIndex{3}\): Finding the tangent line to the folium of Descartes at \(\left(\frac{3}{2},\frac{3}{2}\right)\).

Solution

Begin by implicitly differentiating both sides.\[ \begin{array}{rrcl}
& \dfrac{d}{dx} \left( y^3+x^3−3xy \right) & = & \dfrac{d}{dx} \left( 0 \right) \\[16pt]
\implies & 3y^2\dfrac{dy}{dx}+3x^2−\left(3y+3x\dfrac{dy}{dx}\right) & = & 0 \\[16pt]
\implies & 3y^2\dfrac{dy}{dx}+3x^2−3y-3x\dfrac{dy}{dx} & = & 0 \\[16pt]
\end{array} \nonumber \]It is important to read what we are asked to do now. We are not being asked to solve for \(dy/dx\), but instead to find the equation of the tangent line at \(\left(\frac{3}{2},\frac{3}{2}\right)\). As such, we let \(x = \frac{3}{2}\) and \(y = \frac{3}{2}\) in our implicitly-differentiated equation.\[ \begin{array}{rrcl}
& 3 \left( \dfrac{3}{2} \right)^2 \dfrac{dy}{dx}+3 \left( \dfrac{3}{2} \right)^2−3 \left( \dfrac{3}{2} \right) -3 \left( \dfrac{3}{2} \right) \dfrac{dy}{dx} & = & 0 \\[16pt]
\implies & \dfrac{27}{4} \dfrac{dy}{dx} + \dfrac{27}{4} − \dfrac{9}{2} - \dfrac{9}{2} \dfrac{dy}{dx} & = & 0 \\[16pt]
\implies & 27 \dfrac{dy}{dx} + 27 − 18 - 18 \dfrac{dy}{dx} & = & 0 \\[16pt]
\implies & 9 \dfrac{dy}{dx} + 9 & = & 0 \\[16pt]
\implies & \dfrac{dy}{dx} & = & -1 \\[16pt]
\end{array} \nonumber \]Finally, substitute into the point-slope equation of the line to obtain\[y=−x+3.\nonumber \]

Example \(\PageIndex{6}\): Applying Implicit Differentiation

In a simple video game, a rocket travels in an elliptical orbit whose path is described by the equation \(4x^2+25y^2=100\). The rocket can fire missiles along lines tangent to its path. The object of the game is to destroy an incoming asteroid traveling along the positive \(x\)-axis toward \((0,0)\). If the rocket fires a missile when it is located at \(\left(3,\frac{8}{5}\right)\), where will it intersect the \(x\)-axis?

Solution

To solve this problem, we must determine where the line tangent to the graph of \(4x^2+25y^2=100\) at \(\left(3,\frac{8}{5}\right)\) intersects the \(x\)-axis. Begin by finding \(\frac{dy}{dx}\) implicitly. Differentiating, we have\[8x+50y\dfrac{dy}{dx}=0.\nonumber\]Letting \(x = 3\) and \(y = \frac{8}{5}\), we get\[ 8(3) + 50\left(\dfrac{8}{5}\right) \dfrac{dy}{dx} = 0 \implies 24 + 80 \dfrac{dy}{dx} = 0 \implies \dfrac{dy}{dx} = -\dfrac{3}{10}. \nonumber \]The equation of the tangent line is \(y=−\frac{3}{10}x+\frac{5}{2}\). To determine where the line intersects the \(x\)-axis, solve \(0=−\frac{3}{10}x+\frac{5}{2}\). The solution is \(x=\frac{25}{3}\). The missile intersects the \(x\)-axis at the point \(\left(\frac{25}{3},0\right)\).

Checkpoint \(\PageIndex{6}\)

Find the equation of the line tangent to the hyperbola \(x^2−y^2=16\) at the point \((5,3)\).

Answer

\(y=\frac{5}{3}x−\frac{16}{3}\)

Derivatives of Inverse Trigonometric Functions

We now focus on finding derivatives of inverse trigonometric functions. These derivatives will prove invaluable in the study of integration later in this text. The derivatives of inverse trigonometric functions are quite surprising in that their derivatives are algebraic functions. Previously, derivatives of algebraic functions have proven to be algebraic functions, and derivatives of trigonometric functions have been shown to be trigonometric functions. Here, for the first time, we see that the derivative of a function need not be of the same type as the original function.

Theorem: Derivatives of the Inverse Trigonometric Functions

\[ \begin{array}{rclcrcl}
\dfrac{d}{dx} \left( \sin^{−1}x \right) & = & \dfrac{1}{\sqrt{1−x^2}} & \quad & \dfrac{d}{dx} \left( \csc^{−1}x \right) & = & -\dfrac{1}{x \sqrt{x^2−1}} \\[16pt]
\dfrac{d}{dx} \left( \cos^{−1}x \right) & = & -\dfrac{1}{\sqrt{1−x^2}} & \quad & \dfrac{d}{dx} \left( \sec^{−1}x \right) & = & \dfrac{1}{x \sqrt{x^2−1}} \\[16pt]
\dfrac{d}{dx} \left( \tan^{−1}x \right) & = & \dfrac{1}{1+x^2} & \quad & \dfrac{d}{dx} \left( \cot^{−1}x \right) & = & -\dfrac{1}{1+x^2} \\[16pt]
\end{array} \nonumber \]

Proof that \( \dfrac{d}{dx}{ \left( \sin^{-1}{(x)} \right) } = \dfrac{1}{\sqrt{1 - x^2}} \)

A critical step in this proof is to recall that the ranges of the inverse trigonometric functions are restricted. For the arcsine, we have\[ -\dfrac{\pi}{2} \leq \sin^{-1}{(x)} \leq \dfrac{\pi}{2}. \nonumber \]Therefore,\[ y = \sin^{-1}{(x)}, \text{ where } \sin{(y)} = x \text{ and } -\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}. \nonumber \]An illustration of where the angle \(y\) lies and the right triangles formed from this arcsine can be seen in Figure \(\PageIndex{4}\).

Figure \(\PageIndex{4}\): A graph of where the arcsine returns and the corresponding triangles.

From Figure \(\PageIndex{2}\), we can see that the length of the adjacent side, in either case, is \(+\sqrt{1 - x^2}\) (positive because these triangles both share adjacent sides along the positive \(x\)-axis).

If \(\sin{(y)} = x\), then\[ \begin{array}{rclcrcl}
\dfrac{d}{dx}\left( \sin{(y)} \right) & = & \dfrac{d}{dx}(x) & \implies & \cos{(y)} \dfrac{dy}{dx} & = & 1 \\[16pt]
& & & \implies & \dfrac{dy}{dx} & = & \dfrac{1}{\cos{(y)}} \\[16pt]
& & & \implies & & = & \dfrac{1}{\sqrt{1 - x^2}}. \\[16pt]
\end{array} \nonumber \]

Q.E.D.

Caution: Inverse Secant and Inverse Cosecant Ranges

As was mentioned in Chapter 1 of this textbook, the definition we use for the ranges of the secant and cosecant functions are very specific to this book. A lot of other texts use a different definition. I have chosen to restrict the ranges of the inverse cosecant to \((0,\pi/2] \cup (\pi,3\pi/2]\) and the inverse secant to \([0,\pi/2) \cup [\pi,3\pi/2)\).

This choice is made to avoid challenging issues in Calculus.

The proof we just did indicates the thought process you must go through when proving the derivative of an inverse trigonometric function. You must address the fact that the inverse function has a restricted range. The possible angles an inverse trigonometric function returns are always limited to two quadrants. For each quadrant, drawing a right triangle whose hypotenuse lies on the terminal side of an angle in that restricted quadrant is beneficial.

Example \(\PageIndex{7}\): Finding the Derivative of an Inverse Trigonometric Function

Find the derivative of \(h(x)=\sin^{−1}(2x^3).\)

Solution

\[ \begin{array}{rcl}
\dfrac{d}{dx} \left( \sin^{-1}{(2x^3)} \right) & = & \dfrac{1}{\sqrt{1 - \left(2x^3\right)^2}} \cdot \dfrac{d}{dx}\left( 2x^3\right) \\[16pt]
& = & \dfrac{1}{\sqrt{1 - 4x^6}} \cdot 6x^2 \\[16pt]
& = & \dfrac{6x^2}{\sqrt{1 - 4x^6}} \\[16pt]
\end{array} \nonumber \]

Checkpoint \(\PageIndex{7}\)

Use the inverse function theorem to find the derivative of \(g(x)=\tan^{−1}x\).

Answer

\(g^{\prime}(x)=\frac{1}{1+x^2}\)

Example \(\PageIndex{8}\): Applying Differentiation Formulas to an Inverse Tangent Function

Find the derivative of \(f(x)=\tan^{−1}(x^2).\)

Solution

\[ \begin{array}{rcl}
\dfrac{d}{dx} \left( \tan^{-1}{(x^2)} \right) & = & \dfrac{1}{1 + \left( x^2 \right)^2 } \cdot \dfrac{d}{dx} \left( x^2 \right) \\[16pt]
& = & \dfrac{2x}{1 + x^4} \\[16pt]
\end{array} \nonumber \]

Example \(\PageIndex{9}\): Applying Differentiation Formulas to an Inverse Cosecant Function

Find the derivative of \(h(x)=x^2 \csc^{−1}{(x)}.\)

Solution

\[ \begin{array}{rcl}
\dfrac{d}{dx} \left( x^2 \csc^{-1}{(x)} \right) & = & 2x \csc^{-1}{(x)} - \dfrac{x^2}{x \sqrt{x^2 - 1}} \\[16pt]
& = & 2x \csc^{-1}{(x)} - \dfrac{x}{\sqrt{x^2 - 1}} \\[16pt]
\end{array} \nonumber \]

Checkpoint \(\PageIndex{9}\)

Find the derivative of \(h(x)=\cos^{−1}(3x−1).\)

Answer

\(h^{\prime}(x)=\frac{−3}{\sqrt{6x−9x^2}}\)

Example \(\PageIndex{10}\): Applying the Inverse Tangent Function

The position of a particle at time \(t\) is given by \(s(t)=\tan^{−1}\left(\frac{1}{t}\right)\) for \(t \geq \ce{1/2}\). Find the velocity of the particle at time \( t=1\).

Solution

Begin by differentiating \(s(t)\) in order to find \(v(t)\). Thus,\[v(t)=s^{\prime}(t)=\dfrac{1}{1+\left(\frac{1}{t}\right)^2} \cdot \dfrac{−1}{t^2}.\nonumber\]Simplifying, we have\[v(t)=−\dfrac{1}{t^2+1}.\nonumber\]Thus, \(v(1)=−\frac{1}{2}.\)

Checkpoint \(\PageIndex{10}\)

Find the equation of the line tangent to the graph of \(f(x)=\sin^{−1}x\) at \(x=0.\)

Answer

\(y=x\)

Derivatives of Inverse Hyperbolic Functions

Looking at the graphs of the hyperbolic functions, we see that with appropriate range restrictions, they all have inverses. Most of the necessary range restrictions can be discerned by closely examining the graphs. The domains and ranges of the inverse hyperbolic functions are summarized in Table \(\PageIndex{1}\).

Table \(\PageIndex{1}\): Domains and Ranges of the Inverse Hyperbolic Functions

Function	Domain	Range
\(\sinh^{−1}x\)	\( (−\infty, \infty) \)	\( (− \infty, \infty) \)
\(\cosh^{−1}x\)	\( (1, \infty) \)	\( [0, \infty) \)
\(\tanh^{−1}x\)	\( (−1,1) \)	\( (− \infty, \infty) \)
\(\coth^{−1}x\)	\( (− \infty,1) \cup (1, \infty) \)	\( (− \infty,0) \cup (0, \infty) \)
\(\text{sech}^{−1}x\)	\( (0,1) \)	\( [0, \infty) \)
\(\text{csch}^{−1}x\)	\( (− \infty,0) \cup (0, \infty) \)	\( (− \infty,0) \cup (0, \infty) \)

The graphs of the inverse hyperbolic functions are shown in the following figure.

Figure \(\PageIndex{5}\): Graphs of the inverse hyperbolic functions.

To find the derivatives of the inverse functions, we use implicit differentiation. We have\[ \begin{array}{rcl}
y & = & \sinh^{−1}x\\[16pt]
\sinh y &= & x\\[16pt]
\dfrac{d}{dx} \sinh y & = & \dfrac{d}{dx}x\\[16pt]
\cosh y\dfrac{dy}{dx} & = & 1.\\[16pt]
\end{array} \nonumber \]Recall that \(\cosh^2y−\sinh^2y=1,\) so \(\cosh y=\sqrt{1+\sinh^2y}\).Then,\[\dfrac{dy}{dx}=\dfrac{1}{\cosh y}=\dfrac{1}{\sqrt{1+\sinh^2y}}=\dfrac{1}{\sqrt{1+x^2}}. \nonumber \]We can derive differentiation formulas for the other inverse hyperbolic functions similarly. These differentiation formulas are summarized in Table \(\PageIndex{2}\).

Table \(\PageIndex{2}\): Derivatives of the Inverse Hyperbolic Functions

\(f(x)\)	\(\dfrac{d}{dx}f(x)\)
\(\sinh^{−1}x\)	\(\dfrac{1}{\sqrt{1+x^2}}\)
\(\cosh^{−1}x\)	\(\dfrac{1}{\sqrt{x^2−1}}\)
\(\tanh^{−1}x\)	\(\dfrac{1}{1−x^2}\)
\(\coth^{−1}x\)	\(\dfrac{1}{1−x^2}\)
\(\text{sech}^{−1}x\)	\(\dfrac{−1}{x\sqrt{1−x^2}}\)
\(\text{csch}^{−1}x\)	\(\dfrac{−1}{|x|\sqrt{1+x^2}}\)

Note that the derivatives of \(\tanh^{−1}x\) and \(\coth^{−1}x\) are the same.

Example \(\PageIndex{11}\): Differentiating Inverse Hyperbolic Functions

Evaluate the following derivatives:

1. \(\frac{d}{dx}\left(\sinh^{−1}\left(\frac{x}{3}\right)\right)\)
2. \(\frac{d}{dx}\left(\tanh^{−1}x\right)^2\)

Solution

Using the formulas in Table \(\PageIndex{3}\) and the Chain Rule, we obtain the following results:

1. \(\frac{d}{dx}(\sinh^{−1}(\frac{x}{3}))=\frac{1}{3\sqrt{1+\frac{x^2}{9}}}=\frac{1}{\sqrt{9+x^2}}\)
2. \(\frac{d}{dx}(\tanh^{−1}x)^2=\frac{2(\tanh^{−1}x)}{1−x^2}\)

Checkpoint \(\PageIndex{11}\)

Evaluate the following derivatives:

1. \(\frac{d}{dx}(\cosh^{−1}(3x))\)
2. \(\frac{d}{dx}(\coth^{−1}x)^3\)

Answer a

\(\frac{d}{dx}(\cosh^{−1}(3x))=\frac{3}{\sqrt{9x^2−1}} \)

Answer b

\(\frac{d}{dx}(\coth^{−1}x)^3=\frac{3(\coth^{−1}x)^2}{1−x^2} \)