Setting Up the Inverse Tangent
The tangent function is not one-to-one on its whole domain, but it is strictly increasing on the open interval \(\left(-\tfrac\pi2,\tfrac\pi2\right)\), where it takes every real value exactly once. The arctangent is the inverse of tangent restricted to that interval. For each real \(x\), \(y=\arctan x\) is the unique number in \(\left(-\tfrac\pi2,\tfrac\pi2\right)\) with \(\tan y=x\). The derivative of arctangent follows by differentiating this defining relation. The same method gave the derivatives of arcsine and arccosine earlier in the module.
Derivation from tan y = x
Let \(y=\arctan x\), so \(\tan y=x\) with \(-\tfrac\pi2<y<\tfrac\pi2\). We take for granted that the inverse of a differentiable, strictly monotone function with nonzero derivative is differentiable, so \(y\) is a differentiable function of \(x\).
Differentiating \(\tan y=x\) with respect to \(x\) and using the Chain Rule on the left gives \(\sec^2y\cdot\dfrac{dy}{dx}=1\). Note that \(\sec^2y\ge1>0\), so we may divide.
Divide the identity \(\sin^2y+\cos^2y=1\) by \(\cos^2y\) (which is nonzero because \(y\) is strictly between \(-\tfrac\pi2\) and \(\tfrac\pi2\)). This gives \(\tan^2y+1=\sec^2y\). Since \(\tan y=x\), we get \(\sec^2y=1+x^2\).
Therefore \(\dfrac{dy}{dx}=\dfrac{1}{\sec^2y}=\dfrac{1}{1+x^2}\).
Two features of the result are worth noting. First, the formula holds for every real \(x\); unlike arcsine, arctangent has no points of vertical tangent. Second, the derivative is always positive and is largest at \(x=0\), where it equals \(1\). So the graph of the arctangent rises everywhere, steepest at the origin, and flattens out toward the horizontal asymptotes \(y=\pm\tfrac\pi2\) as \(|x|\) grows.
Worked Example: Checking the Formula on Sample Inputs
Verify \(\dfrac{d}{dx}\arctan x=\dfrac1{1+x^2}\) at \(x=0\), \(x=1\) and \(x=\sqrt3\).
The formula predicts the values
Compare with difference quotients using known values of arctangent: \(\arctan0=0\), \(\arctan1=\tfrac\pi4\approx0.785398\), and \(\arctan\sqrt3=\tfrac\pi3\approx1.047198\). Near \(x=1\), \(\arctan1.001\approx0.785898\) and \(\arctan0.999\approx0.784898\), so the central difference quotient is \(\dfrac{0.785898-0.784898}{0.002}=0.5000\), matching \(\tfrac12\). The same estimate at the other two points also agrees with the predicted values.
Worked Example: A Linear Inner Function
Differentiate \(f(x)=\arctan(3x)\).
With \(u=3x\) and \(u'=3\), the Chain Rule gives
Check at \(x=1\): the formula gives \(\dfrac3{10}=0.3\). A difference quotient with \(\arctan3.003\approx1.249346\) and \(\arctan2.997\approx1.248746\) gives \(\dfrac{1.249346-1.248746}{0.002}=0.3000\).
Worked Example: Reciprocal Argument
Differentiate \(g(x)=\arctan\left(\dfrac1x\right)\) for \(x\ne0\).
Take \(u=1/x\), so \(u'=-1/x^2\). Then
At \(x=2\), \(g'(2)=-\tfrac15\). This result has an interpretation. For \(x>0\), \(\arctan x+\arctan(1/x)=\tfrac\pi2\), which is constant, so the derivative of \(\arctan(1/x)\) must be the negative of the derivative of \(\arctan x\). Indeed \(-\dfrac1{1+x^2}\) is exactly that negative.
Worked Example: A Tangent Line
Find the tangent line to \(y=\arctan x\) at \(x=1\).
The point is \((1,\tfrac\pi4)\) and the slope is \(\dfrac{1}{1+1}=\tfrac12\). The tangent line is
At \(x=1.2\) the line predicts \(y\approx0.785398+0.1=0.885398\), and the true value is \(\arctan1.2\approx0.876058\), a small difference as expected for a short step.
Common Mistakes
Confusing the inverse with the reciprocal. The notation \(\tan^{-1}x\) means \(\arctan x\), not \(\dfrac{1}{\tan x}\). The derivative of \(\cot x=\dfrac1{\tan x}\) is \(-\csc^2x\), which is a different function entirely.
Leaving the answer in terms of \(y\). The derivation produces \(\dfrac{1}{\sec^2y}\). The result must be rewritten as a function of \(x\), which is why the identity \(\sec^2y=1+\tan^2y\) is needed.
Dropping the Chain Rule factor. The derivative of \(\arctan(3x)\) is \(\dfrac3{1+9x^2}\), not \(\dfrac{1}{1+9x^2}\). At \(x=0\) the correct slope is \(3\), reflecting that the argument \(3x\) changes three times as fast as \(x\).
Worked Example: Arctangent Inside a Product
Differentiate \(p(x)=x\arctan x\) and evaluate \(p'(1)\).
Use the Product Rule with \(\dfrac{d}{dx}x=1\) and \(\dfrac{d}{dx}\arctan x=\dfrac1{1+x^2}\):
At \(x=1\), \(p'(1)=\dfrac\pi4+\dfrac12\approx0.785398+0.5=1.285398\).
Check: with \(p(x)=x\arctan x\), \(p(1.001)\approx0.786684\) and \(p(0.999)\approx0.784113\), giving a difference quotient of about \(1.285\).
Summary of the Derivative Formulas
| Function | Derivative | Domain of validity |
|---|---|---|
| \(\arctan x\) | \(\dfrac{1}{1+x^2}\) | all real \(x\) |
| \(\arctan u(x)\) | \(\dfrac{u'(x)}{1+u(x)^2}\) | \(u\) differentiable |
| \(\arctan(ax)\) | \(\dfrac{a}{1+a^2x^2}\) | all real \(x\), \(a\) constant |
Notice that every denominator has the form \(1+(\text{argument})^2\) and every numerator is the derivative of the argument. This is the pattern to look for when you meet such a derivative in a problem, and when you meet the reverse pattern in an integral. If you forget the formula, you can rebuild it in four lines from \(\tan y=x\), which is far more reliable than recalling a table.
Worked Example: Concavity of the Arctangent
Find the second derivative of \(\arctan x\) and describe its concavity.
Write \(\dfrac{d}{dx}\arctan x=(1+x^2)^{-1}\) and differentiate with the Chain Rule:
The denominator is positive, so the sign of the second derivative is the sign of \(-2x\). The arctangent is concave up for \(x<0\), concave down for \(x>0\), and has an inflection point at the origin where its slope \(1\) is largest.
Check: at \(x=1\), the second derivative is \(-\dfrac{2}{4}=-0.5\), which equals the derivative of \(\dfrac{1}{1+x^2}\) at \(1\), namely \(-\dfrac{2x}{(1+x^2)^2}=-0.5\). The two expressions agree because they are the same quantity.
Worked Example: A Linear Approximation Near the Origin
Estimate \(\arctan(0.1)\) using the tangent line at \(x=0\).
At \(x=0\), \(\arctan0=0\) and the slope is \(1\), so the linearization is \(L(x)=x\). Thus \(\arctan(0.1)\approx0.1\).
Check: the true value is \(\arctan0.1\approx0.0996687\), so the estimate is high by about \(0.000331\). Because the curve is concave down for \(x>0\), the tangent line lies above it, so the overestimate is the expected direction.
Notice how this tutorial used the same pattern as the other inverse-function derivatives. In each case, the inverse relation is written in a form that can be differentiated, the derivative of the forward function appears, and the Pythagorean-type identity converts the result back to \(x\). With arctangent, the identity \(\sec^2y=1+\tan^2y\) makes the answer algebraic: there is no square root, unlike the derivatives of arcsine and arccosine, which involve \(\sqrt{1-x^2}\). That algebraic simplicity is why the arctangent appears so often in integrals of rational functions.
Worked Example: A Composition with Exponential and Arctangent
Differentiate \(q(x)=\arctan(e^{x})\) and find the limit of \(q'(x)\) as \(x\to\infty\) and as \(x\to-\infty\).
With \(u=e^x\) and \(u'=e^x\), the formula gives \(q'(x)=\dfrac{e^x}{1+e^{2x}}\). Dividing the numerator and the denominator by \(e^x\) gives the equivalent form \(\dfrac{1}{e^{-x}+e^{x}}=\dfrac{1}{2\cosh x}\).
As \(x\to\pm\infty\), the denominator \(e^{-x}+e^{x}\) grows without bound, so \(q'(x)\to0\) in both directions. At \(x=0\), \(q'(0)=\dfrac{1}{2}\), the largest value. So \(q\) increases everywhere but flattens out toward its horizontal asymptotes \(\arctan(0)=0\) as \(x\to-\infty\) and \(\tfrac\pi2\) as \(x\to\infty\).
Where This Fits in the Course
This tutorial belongs to the module “Implicit and Inverse Differentiation”. The tutorial just before it is “Deriving The Arccosine Derivative,” and the one that follows is “Deriving The Arccotangent Derivative.”
Using the Result for Antiderivatives
Because \(\dfrac{d}{dx}\arctan x=\dfrac{1}{1+x^2}\), the function \(\arctan x\) is an antiderivative of \(\dfrac{1}{1+x^2}\), which shows that \(\displaystyle\int\frac{dx}{1+x^2}=\arctan x+C\). Definite integrals follow at once: \(\displaystyle\int_0^1\frac{dx}{1+x^2}=\arctan1-\arctan0=\frac\pi4\). The Chain Rule example \(\dfrac{d}{dx}\arctan(3x)=\dfrac{3}{1+9x^2}\) likewise gives \(\displaystyle\int\frac{3\,dx}{1+9x^2}=\arctan(3x)+C\). These antiderivatives are the main reason the arctangent derivative matters in later work.
Practice Problems with Answers
Try each problem before reading the answer. Each answer below was verified by differentiation or by evaluating a difference quotient or integral numerically.
- Differentiate \(\arctan(x^2)\) and evaluate at \(x=1\).
Answer. \(\dfrac{2x}{1+x^4}\); at \(x=1\) it equals \(1\). - Differentiate \(x\arctan x-\tfrac12\ln(1+x^2)\).
Answer. \(\arctan x+\dfrac{x}{1+x^2}-\dfrac{x}{1+x^2}=\arctan x\). - Differentiate \(\arctan(e^x)\).
Answer. \(\dfrac{e^x}{1+e^{2x}}\). - Find the equation of the tangent line to \(y=\arctan x\) at \(x=0\).
Answer. The slope is \(1\) and \(\arctan0=0\), so \(y=x\).
Check Your Understanding
Work these without looking back.
- Differentiate \(\arctan(x^2)\) and evaluate the derivative at \(x=1\).
- Differentiate \(x\arctan x-\tfrac12\ln(1+x^2)\) and simplify.
- Why is \(\sec^2y\ne0\) in the derivation, and where is that fact used?
- Find the limit of \(\arctan'(x)\) as \(x\to\infty\) and interpret it.
- Show that \(\arctan x+\arctan(1/x)\) has derivative \(0\) for \(x>0\).
