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I have seen the derivative of $\tan^{-1}(x)$ set equal to both $\sec^{-2}(x)$ and $\frac{1}{1+x^2}$.

I can't find corroborating sources for either claim, so I'm asking here if either of these are true.

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6 Answers

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This stems from the fact that many sources use the conflicting definitions that$$\tan^{-1}{(x)}=\arctan{(x)}$$But one may also interpret this as$$\tan^{-1}{(x)}=\frac1{\tan{(x)}}$$Both of these functions have different derivatives.

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It boils down to: what do you mean by $\tan^{-1}(x)$. I think most people think of it as the inverse tangent function, i.e. $\arctan(x)$, but some think of it as $\frac{1}{\tan(x)}$. The derivative of the former is $\frac{1}{1+x^2}$, and the derivative of $\tan(x)$ is $\sec^2(x)$.

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If by $\tan^{-1}$ you mean the inverse function of the restriction of $\tan$ to the interval $(-\pi/2,\pi/2)$, i.e. the function $\arctan$, you can apply the general formula for the derivative of an inverse function:$$(\arctan)'(x)=\frac 1{(\tan)'(\arctan x)}==\frac 1{1+\tan^2(\arctan x)}=\frac 1{1+x^2}.$$

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If $y=\arctan x$ then $x=\tan y$ so $\frac{dx}{dy}=\sec^2y=1+x^2$. Therefore, $\frac{dy}{dx}$ is $\sec^{-2}y$ (note it's $y$, not $x$; also, we'd usually write $\cos^2 y$) or $\frac{1}{1+x^2}$.

On the other hand, if $y=(\tan x)^{-1}$ then $\frac{dy}{dx}=-\tan^{-2}x\sec^2 x=-\csc^2 x=-(1+y^2)$.

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The derivative of the inverse tangent function ($f(x)=\tan^{-1}{x}$), also commonly known as the arctangent function ($f(x)=\arctan{x}$), is: $$\frac{1}{1+x^2}.$$

Period. It's definitely not $\sec^{-2}{x}$. That's just pure nonsense. In fact, if you are thinking of $\tan^{-1}{x}$ as the reciprocal of the tangent function, then the derivative of $\frac{1}{\tan{x}}$ would actually be $-\csc^2{x}$:

$$ \frac{d}{dx}\left(\frac{1}{\tan{x}}\right)=\frac{d}{dx}\bigg[(\tan{x})^{-1}\bigg]=-1\cdot\left(\tan{x}\right)^{-1-1}\frac{d}{dx}\left(\tan{x}\right)=\\ -\frac{1}{\tan^2{x}}\cdot\sec^2{x}=-\frac{1}{\frac{\sin^2{x}}{\cos^2{x}}}\cdot\frac{1}{\cos^2{x}}=-\frac{1}{\sin^2{x}}=-\csc^2{x}. $$

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We have $\tan(\arctan(x)) = x$, and we can differentiate implicitly:

$$\tan'(\arctan(x))\arctan'(x) = 1$$

Rearranging:

$$\arctan'(x) = 1/\sec^2(\arctan(x))$$

See that $\sec(\arctan(x)) = \sqrt{1 + x^2}$ from the right triangle:

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So $\displaystyle \frac{\text{d}}{\text{d}x} \arctan(x) = \frac{1}{1 + x^2}$.

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Note: This approach can be applied to find the derivative of any inverse function, assuming you know the derivative of the function itself. To wit:

Suppose $g(x) = f^{-1}(x)$. Then:

\begin{align} &f(g(x)) = x \\[1em] &f'(g(x)) \cdot g'(x) = 1 \qquad \quad \text{(Differentiating implicitly)} \\[1em] &g'(x) = \frac{1}{f'(g(x))} \end{align}

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