Derivative of arctan(x) Let’s use our formula for the derivative of an inverse function to find the deriva tive of the inverse of the tangent function: y = tan−1 x = arctan x. We simplify the equation by taking the tangent of both sides: y = tan−1 x tan y = tan(tan−1 x) tan y = x. Several notations for the inverse trigonometric functions exist. The most common convention is to name inverse trigonometric functions using an arc- prefix: arcsin(x), arccos(x), arctan(x), etc. (This convention is used throughout this article.) This notation arises from the following geometric relationships: citation needed when measuring in radians, an angle of θ.
Learning Objectives
- Calculate the derivative of an inverse function.
- Recognize the derivatives of the standard inverse trigonometric functions.
In this section we explore the relationship between the derivative of a function and the derivative of its inverse. For functions whose derivatives we already know, we can use this relationship to find derivatives of inverses without having to use the limit definition of the derivative. In particular, we will apply the formula for derivatives of inverse functions to trigonometric functions. This formula may also be used to extend the power rule to rational exponents.
The Derivative of an Inverse Function
We begin by considering a function and its inverse. If (f(x)) is both invertible and differentiable, it seems reasonable that the inverse of (f(x)) is also differentiable. Figure (PageIndex{1}) shows the relationship between a function (f(x)) and its inverse (f^{−1}(x)). Look at the point (left(a,f^{−1}(a)right)) on the graph of (f^{−1}(x)) having a tangent line with a slope of
[big(f^{−1}big)′(a)=dfrac{p}{q}.]
This point corresponds to a point (left(f^{−1}(a),aright)) on the graph of (f(x)) having a tangent line with a slope of
[f′big(f^{−1}(a)big)=dfrac{q}{p}.]
Thus, if (f^{−1}(x)) is differentiable at (a), then it must be the case that
(big(f^{−1}big)′(a)=dfrac{1}{f′big(f^{−1}(a)big)}).
We may also derive the formula for the derivative of the inverse by first recalling that (x=fbig(f^{−1}(x)big)). Then by differentiating both sides of this equation (using the chain rule on the right), we obtain
(1=f′big(f^{−1}(x)big)big(f^{−1}big)′(x))).
Solving for (big(f^{−1}big)′(x)), we obtain
(big(f^{−1}big)′(x)=dfrac{1}{f′big(f^{−1}(x)big)}).
We summarize this result in the following theorem.
Inverse Function Theorem
Let (f(x)) be a function that is both invertible and differentiable. Let (y=f^{−1}(x)) be the inverse of (f(x)). For all (x) satisfying (f′big(f^{−1}(x)big)≠0),
[dfrac{dy}{dx}=dfrac{d}{dx}big(f^{−1}(x)big)=big(f^{−1}big)′(x)=dfrac{1}{f′big(f^{−1}(x)big)}.label{inverse1}]
Alternatively, if (y=g(x)) is the inverse of (f(x)), then
[g'(x)=dfrac{1}{f′big(g(x)big)}. label{inverse2}]
Example (PageIndex{1}): Applying the Inverse Function Theorem
Use the inverse function theorem to find the derivative of (g(x)=dfrac{x+2}{x}). Compare the resulting derivative to that obtained by differentiating the function directly.
Solution
The inverse of (g(x)=dfrac{x+2}{x}) is (f(x)=dfrac{2}{x−1}).
We will use Equation ref{inverse2} and begin by finding (f′(x)). Thus,
[f′(x)=dfrac{−2}{(x−1)^2} nonumber ]
and
[f′big(g(x)big)=dfrac{−2}{(g(x)−1)^2}=dfrac{−2}{left(dfrac{x+2}{x}−1right)^2}=−dfrac{x^2}{2}. nonumber ]
Finally,
[g′(x)=dfrac{1}{f′big(g(x)big)}=−dfrac{2}{x^2}. nonumber ]
We can verify that this is the correct derivative by applying the quotient rule to (g(x)) to obtain
[g′(x)=−dfrac{2}{x^2}. nonumber ]
Exercise (PageIndex{1})
Use the inverse function theorem to find the derivative of (g(x)=dfrac{1}{x+2}). Compare the result obtained by differentiating (g(x)) directly.
Use the preceding example as a guide.
(g′(x)=−dfrac{1}{(x+2)^2})
Example (PageIndex{2}): Applying the Inverse Function Theorem
Use the inverse function theorem to find the derivative of (g(x)=sqrt[3]{x}).
Solution
The function (g(x)=sqrt[3]{x}) is the inverse of the function (f(x)=x^3). Since (g′(x)=dfrac{1}{f′big(g(x)big)}), begin by finding (f′(x)). Thus,
[f′(x)=3x^3nonumber]
and
[f′big(g(x)big)=3big(sqrt[3]{x}big)^2=3x^{2/3}nonumber]
Finally,
[g′(x)=dfrac{1}{3x^{2/3}}.nonumber]
If we were to integrate (g(x)) directing, using the power rule, we would first rewrite (g(x)=sqrt[3]{x}) as a power of (x) to get,
[g(x) = x^{1/3}nonumber]
Then we would differentiate using the power rule to obtain
[g'(x) =tfrac{1}{3}x^{−2/3} = dfrac{1}{3x^{2/3}}.nonumber]
Exercise (PageIndex{2})
Find the derivative of (g(x)=sqrt[5]{x}) by applying the inverse function theorem.
(g(x)) is the inverse of (f(x)=x^5).
(g(x)=frac{1}{5}x^{−4/5})
From the previous example, we see that we can use the inverse function theorem to extend the power rule to exponents of the form (dfrac{1}{n}), where (n) is a positive integer. This extension will ultimately allow us to differentiate (x^q), where (q) is any rational number.
Extending the Power Rule to Rational Exponents
The power rule may be extended to rational exponents. That is, if (n) is a positive integer, then
[dfrac{d}{dx}big(x^{1/n}big)=dfrac{1}{n} x^{(1/n)−1}.]
Also, if (n) is a positive integer and (m) is an arbitrary integer, then
[dfrac{d}{dx}big(x^{m/n}big)=dfrac{m}{n}x^{(m/n)−1}.]
Proof
The function (g(x)=x^{1/n}) is the inverse of the function (f(x)=x^n). Since (g′(x)=dfrac{1}{f′big(g(x)big)}), begin by finding (f′(x)). Thus,
(f′(x)=nx^{n−1}) and (f′big(g(x)big)=nbig(x^{1/n}big)^{n−1}=nx^{(n−1)/n}).
Finally,
(g′(x)=dfrac{1}{nx^{(n−1)/n}}=dfrac{1}{n}x^{(1−n)/n}=dfrac{1}{n}x^{(1/n)−1}).
To differentiate (x^{m/n}) we must rewrite it as ((x^{1/n})^m) and apply the chain rule. Thus,
[dfrac{d}{dx}big(x^{m/n}big)=dfrac{d}{dx}big((x^{1/n}big)^m)=mbig(x^{1/n}big)^{m−1}⋅dfrac{1}{n}x^{(1/n)−1}=dfrac{m}{n}x^{(m/n)−1}. nonumber]
□
Example (PageIndex{3}): Applying the Power Rule to a Rational Power
Find the equation of the line tangent to the graph of (y=x^{2/3}) at (x=8).
Solution
First find (dfrac{dy}{dx}) and evaluate it at (x=8). Since
[dfrac{dy}{dx}=frac{2}{3}x^{−1/3} nonumber]
and
[dfrac{dy}{dx}Bigg|_{x=8}=frac{1}{3}nonumber ]
the slope of the tangent line to the graph at (x=8) is (frac{1}{3}).
Substituting (x=8) into the original function, we obtain (y=4). Thus, the tangent line passes through the point ((8,4)). Substituting into the point-slope formula for a line, we obtain the tangent line
[y=tfrac{1}{3}x+tfrac{4}{3}. nonumber]
Exercise (PageIndex{3})
Find the derivative of (s(t)=sqrt{2t+1}).
Use the chain rule.
(s′(t)=(2t+1)^{−1/2})
Derivatives of Inverse Trigonometric Functions
We now turn our attention to 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 actually 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.
Example (PageIndex{4A}): Derivative of the Inverse Sine Function
Use the inverse function theorem to find the derivative of (g(x)=sin^{−1}x).
Solution
Since for (x) in the interval (left[−frac{π}{2},frac{π}{2}right],f(x)=sin x) is the inverse of (g(x)=sin^{−1}x), begin by finding (f′(x)). Since
[f′(x)=cos x nonumber]
and
[f′big(g(x)big)=cos big( sin^{−1}xbig)=sqrt{1−x^2} nonumber]
we see that
[g′(x)=dfrac{d}{dx}big(sin^{−1}xbig)=dfrac{1}{f′big(g(x)big)}=dfrac{1}{sqrt{1−x^2}} nonumber]
Analysis
To see that (cos(sin^{−1}x)=sqrt{1−x^2}), consider the following argument. Set (sin^{−1}x=θ). In this case, (sin θ=x) where (−frac{π}{2}≤θ≤frac{π}{2}). We begin by considering the case where (0<θ<frac{π}{2}). Since (θ) is an acute angle, we may construct a right triangle having acute angle (θ), a hypotenuse of length (1) and the side opposite angle (θ) having length (x). From the Pythagorean theorem, the side adjacent to angle (θ) has length (sqrt{1−x^2}). This triangle is shown in Figure (PageIndex{2}) Using the triangle, we see that (cos(sin^{−1}x)=cos θ=sqrt{1−x^2}).
In the case where (−frac{π}{2}<θ<0), we make the observation that (0<−θ<frac{π}{2}) and hence
(cosbig(sin^{−1}xbig)=cos θ=cos(−θ)=sqrt{1−x^2}).
Now if (θ=frac{π}{2}) or (θ=−frac{π}{2},x=1) or (x=−1), and since in either case (cosθ=0) and (sqrt{1−x^2}=0), we have
(cosbig(sin^{−1}xbig)=cosθ=sqrt{1−x^2}).
Consequently, in all cases,
[cosbig(sin^{−1}xbig)=sqrt{1−x^2}.nonumber]
Example (PageIndex{4B}): Applying the Chain Rule to the Inverse Sine Function
Apply the chain rule to the formula derived in Example (PageIndex{4A}) to find the derivative of (h(x)=sin^{−1}big(g(x)big)) and use this result to find the derivative of (h(x)=sin^{−1}(2x^3).)
Solution
Applying the chain rule to (h(x)=sin^{−1}big(g(x)big)), we have
(h′(x)=dfrac{1}{sqrt{1−big(g(x)big)^2}}g′(x)).
Now let (g(x)=2x^3,) so (g′(x)=6x^2). Substituting into the previous result, we obtain
(begin{align*} h′(x)&=dfrac{1}{sqrt{1−4x^6}}⋅6x^2[4pt]&=dfrac{6x^2}{sqrt{1−4x^6}}end{align*})
Exercise (PageIndex{4})
Use the inverse function theorem to find the derivative of (g(x)=tan^{−1}x).
The inverse of (g(x)) is (f(x)=tan x). Use Example (PageIndex{4A}) as a guide.
(g′(x)=dfrac{1}{1+x^2})
The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem. These formulas are provided in the following theorem.
Derivatives of Inverse Trigonometric Functions
[begin{align} dfrac{d}{dx}big(sin^{−1}xbig) &=dfrac{1}{sqrt{1−x^2}} label{trig1} [4pt] dfrac{d}{dx}big(cos^{−1}xbig) &=dfrac{−1}{sqrt{1−x^2}} label{trig2} [4pt] dfrac{d}{dx}big(tan^{−1}xbig) &=dfrac{1}{1+x^2} label{trig3} [4pt] dfrac{d}{dx}big(cot^{−1}xbig) &=dfrac{−1}{1+x^2} label{trig4} [4pt] dfrac{d}{dx}big(sec^{−1}xbig) &=dfrac{1}{|x|sqrt{x^2−1}} label{trig5} [4pt] dfrac{d}{dx}big(csc^{−1}xbig) &=dfrac{−1}{|x|sqrt{x^2−1}} label{trig6} end{align}]
Example (PageIndex{5A}): Applying Differentiation Formulas to an Inverse Tangent Function
Find the derivative of (f(x)=tan^{−1}(x^2).)
Solution
Let (g(x)=x^2), so (g′(x)=2x). Substituting into Equation ref{trig3}, we obtain
(f′(x)=dfrac{1}{1+(x^2)^2}⋅(2x).)
Simplifying, we have
(f′(x)=dfrac{2x}{1+x^4}).
Example (PageIndex{5B}): Applying Differentiation Formulas to an Inverse Sine Function
Find the derivative of (h(x)=x^2 sin^{−1}x.)
Solution
By applying the product rule, we have
(h′(x)=2xsin^{−1}x+dfrac{1}{sqrt{1−x^2}}⋅x^2)
Exercise (PageIndex{5})
Find the derivative of (h(x)=cos^{−1}(3x−1).)
Use Equation ref{trig2}. with (g(x)=3x−1)
(h′(x)=dfrac{−3}{sqrt{6x−9x^2}})
Example (PageIndex{6}): 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≥ 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′(t)=dfrac{1}{1+left(frac{1}{t}right)^2}⋅dfrac{−1}{t^2}).
Simplifying, we have
(v(t)=−dfrac{1}{t^2+1}).
Thus, (v(1)=−dfrac{1}{2}.)
Exercise (PageIndex{6})
Find the equation of the line tangent to the graph of (f(x)=sin^{−1}x) at (x=0.)
(f′(0)) is the slope of the tangent line.
(y=x)
Key Concepts
- The inverse function theorem allows us to compute derivatives of inverse functions without using the limit definition of the derivative.
- We can use the inverse function theorem to develop differentiation formulas for the inverse trigonometric functions.
Key Equations
- Inverse function theorem
((f−1)′(x)=dfrac{1}{f′big(f^{−1}(x)big)}) whenever (f′big(f^{−1}(x)big)≠0) and (f(x)) is differentiable.
- Power rule with rational exponents
(dfrac{d}{dx}big(x^{m/n}big)=dfrac{m}{n}x^{(m/n)−1}.)
- Derivative of inverse sine function
(dfrac{d}{dx}big(sin^{−1}xbig)=dfrac{1}{sqrt{1−x^2}})
- Derivative of inverse cosine function
(dfrac{d}{dx}big(cos^{−1}xbig)=dfrac{−1}{sqrt{1−x^2}})
Derivative of inverse tangent function
(dfrac{d}{dx}big(tan^{−1}xbig)=dfrac{1}{1+x^2})
Derivative of inverse cotangent function
(dfrac{d}{dx}big(cot^{−1}xbig)=dfrac{−1}{1+x^2})
Derivative of inverse secant function
(dfrac{d}{dx}big(sec^{−1}xbig)=dfrac{1}{|x|sqrt{x^2−1}})
Derivative of inverse cosecant function
(dfrac{d}{dx}big(csc^{−1}xbig)=dfrac{−1}{|x|sqrt{x^2−1}})
Contributors and Attributions
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.
- Paul Seeburger (Monroe Community College) added the second half of Example (PageIndex{2}).
DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS
Differentiation of inverse trigonometric functions is a small and specialized topic. However, these particular derivatives are interesting to us for two reasons. First, computation of these derivatives provides a good workout in the use of the chain rule, the definition of inverse functions, and some basic trigonometry. Second, it turns out that the derivatives of the inverse trigonometric functions are actually algebraic functions!! This is an unexpected and interesting connection between two seemingly very different classes of functions.
It is possible to form inverse functions for restricted versions of all six basic trigonometric functions. One can construct and use an inverse cosecant function, for example. However, it is generally enought to consider the inverse sine and the inverse tangent functions. We will restrict our attention to these two functions.
Here are the results.
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Derivative Of Arctan X
In order to verify the differentiation formula for the arcsine function, let us set
y = arcsin(x).We want to compute dy/dx. The first step is to use the fact that the arcsine function is the inverse of the sine function. Among other things, this means that
sin(y) = sin(arcsin(x)) = x.Next, differentiate both ends of this formula. We apply the chain rule to the left end, remembering that the derivative of the sine function is the cosine function and that y is a differentiable function of x.
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The next step is to solve for dy/dx. (After all, this is the thing that we want to compute!)
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| = | sec(y) |
This looks like progress, but it is not the answer. Remember, when we differentiate a function of x in terms of x (this is the meaning of the dx in d/dx), we must express our answer in terms of x. Therefore the question remains.
If sin(y)=x, what is cos(y) in terms of x?
The length of the third side of the reference triangle is determined by the Pythagorean Theorem.
12=x2+The last step is to express the trigonometric functions of y in terms of ratios of side lengths in the reference triangle. In particular, the secant of y is equal to the hypotenuse length divided by the adjacent side length.(third side)2 (third side)2 = 1-x2third side =sqrt(1-x2)
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Derivative Of Arctan X
This completes our study of differentiation for now. In Stage 6, we will investigate another general differentiation technique called implicit differentiation. Later still, we will learn how to differentiate exponential and logarithmic functions. For now, you should go to the Practice area and spend some time learning to use the many differentiation techniques that have been introduced in this Lesson. If there is one skill that we must develop for success in differential calculus, it is differentiation!! Enjoy, and good luck!!
If you find that you are having difficulty with differentiation, don't worry. You're not the first person to struggle with this technical skill. Contact your classmates. Discuss your difficulties. Contact your instructor. We can learn to do this!!
We now turn to the exponential and logarithmic functions. We have already discussed 'the' exponential function: exp(x)=ex. In order to differentiate the 'other' exponential functions and the logarithmic functions, we must first compute the derivative of the inverse to the exponential function. Thus we turn our attention to the natural logarithm.
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Derivative Of Arctan(4x)
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