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My textbook says the antiderivative of $\frac{1}{1+x^2}$ is $\tan^{-1}(x)$.

To confirm this to myself I took the derivative of $\tan^{-1}(x)$ expecting to get $\frac{1}{1+x^2}$ , but instead I ended up with $-\frac{1}{\sin^2(x)}$. So why is $\tan^{-1}(x)$ the antiderivative of $\frac{1}{1+x^2}$ if the derivative of $\tan^{-1}(x)$ is not $\frac{1}{1+x^2}$? Shouldn't the derivative of the antiderivative of a function give you the original function?

k170
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Asker
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6 Answers6

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$$ \begin{align} \arctan(x) &= y\\ x &= \tan(y)\\ \frac{\mathrm d}{\mathrm dx} x &= \frac{\mathrm d}{\mathrm dx} \tan(y)\\ 1 &= y' \sec^2(y)\\ y'&=\dfrac{1}{\sec^2(y)}\\ y'&=\dfrac{1}{\tan^2(y)+1}\\ y'&=\dfrac{1}{x^2 + 1} \end{align} $$

Because $x = \tan(y)$

Nick
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Teoc
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$$x=\arctan y$$ $$y=\tan x$$ $$\frac {dy}{dx}=\sec^2x=1+\tan^2 x=1+y^2$$

$$\frac {dx}{dy}=\frac {1}{1+y^2}$$

Aditya Hase
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$$ y = \tan x $$ $$ \frac {dy}{dx} = \sec^2 x = 1+\tan^2 x = 1+y^2 $$ $$ \frac{dy}{dx} = 1+y^2 $$ $$ \frac{dx}{dy} = \frac{1}{1+y^2} $$ $$ \frac d{dx} \arctan y = \frac 1 {1+y^2} $$

Is the reciprocal of $dy/dx$ really $dx/dy$? It certainly would be if $dy$ and $dx$ were actual numbers. The fact that these are each others reciprocal's is an instance of the chain rule.

Michael Hardy
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  • "The fact that these are each others reciprocal's is an instance of the chain rule." <- Why? – Nick Nov 14 '14 at 03:38
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    @Nick : The chain rule can be written as $\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}$. Apply that to the case where $x$ and $y$ are the same variable: $\dfrac{dx}{dx} = \dfrac{dx}{du}\cdot\dfrac{du}{dx}$, and remember that $dx/dx=1$. ${}\qquad{}$ – Michael Hardy Nov 14 '14 at 03:40
  • So, you're using the fact that $a\cdot b = 1$ implies that $a$ is the multiplicative inverse of $b$ and vice versa. Thanks for this explanation :D – Nick Nov 14 '14 at 03:51
  • $\dfrac{\Delta y}{\Delta x}$ and $\dfrac{\Delta x}{\Delta y}$ are reciprocals (because those are real numbers). (P.S. $\Delta y$ is defined as $y(x+\Delta x)-y(x)$.) Thus, by taking the limit as $\Delta x\to0$, we have that $\dfrac{dy}{dx}$ and $\dfrac{dx}{dy}$ are reciprocals. – Akiva Weinberger Nov 14 '14 at 04:23
  • @columbus8myhw : That's almost right. A problem arises when the set of values of $\Delta x$ for which $\Delta y=0$ is not bounded away from $\Delta x=0$. That gives you zeros in a denominator. But that problem is surmountable by a bit of additional argument. ${}\qquad{}$ – Michael Hardy Nov 14 '14 at 04:35
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Consider the following. $$ x=\tan(y)$$ $$dx = \sec^2(y)dy$$

Therefore.

$$\int\frac{1}{1+x^2}dx=\int\frac{\sec^2(y)}{1+\tan^2(y)}dy$$ $$=\int\frac{\sec^2(y)}{\sec^2(y)}dy$$ $$=\int dy$$ $$=y+c$$ $$=\arctan(x)+c$$

math424
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Using the formula for geometric series, $${1\over 1+x^2}=1-x^2+x^4-x^6+\cdots+(-x^2)^n+\cdots $$ Integrating, $$ \begin{align} \int 1-x^2+x^4-x^6+\cdots+(-x^2)^n+\cdots dx&=x-{x^3\over3}+{x^5\over5}-{x^7\over7}+\cdots\\ &=\arctan x\\ \end{align} $$

Suzu Hirose
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Generally speaking,$f^{-1}$ is the inverse of $f(x)$, not $\frac{1}{f(x)}$.

For example, $\tan^{-1}x=\arctan x$, $\sin^{-1}x=\arcsin x$.

Paul
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    Although both $f^{-1}$ and $\frac 1 f$ are inverses - just different types. One's a function inverse, and ones a multiplicative inverse. – GFauxPas Nov 14 '14 at 03:17
  • $\dfrac1{f(x)}$ is indeed the inverse of $f(x)$, if "inverse" means multiplicative inverse. But this is really about the (compositional) inverse of $f$, not of $f(x)$, but then evaluated at $x$. ${}\qquad{}$ – Michael Hardy Nov 14 '14 at 03:43