I'd like to know how to derive these functions (I know the answers, I want to know how to get there) \begin{align*} f(x) &= \arcsin\left(\frac{x}{3}\right)\\ f(x) &= \arccos(2x+1)\\ f(x) &= \arctan(x^2)\\ f(x) &= \mathrm{arcsec}(x^7)\\ \end{align*} etc.
Trying to derive an inverse trigonometric function
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1The general way is to know the derivatives of the inverse trigonometric functions, and then apply the Chain Rule. You *could* try to derive the derivative of, say $f(x)=\mathrm{arcsec}(x^7)$ by finding the derivative of $\sec(f(x))$, applying the Chain Rule, and then solving for $f'(x)$, but in the long run it is probably going to be less work to remember the derivatives of the inverse trig functions than to derive them from scratch every time you need them. – 2011-02-21
4 Answers
The formulas you need are the derivatives of $\arcsin(u)$, $\arccos(u)$, $\arctan(u)$, $\mathrm{arcsec}(u)$, and presumably $\mathrm{arccot}(u)$ and $\mathrm{arccsc}(u)$. Once you know these, you can apply the Chain Rule.
And how do you find these derivatives? Well, the Inverse Function Theorem is your first friend. If $y = g(x)$ has an inverse, is differentiable at $x=a$, $g(a)=b$, and g'(a)\neq 0, then (g^{-1})'(b) = \frac{1}{g'(a)}.
So, consider $y=\sin(\theta)$. Since the derivative of $\sin(\theta)$ is $\cos(\theta)$, you have that $\frac{d}{du}\arcsin(u) = \frac{1}{\cos(\arcsin(u))}.$ But... what is $\cos(\arcsin(u))$? Suppose $\arcsin(u)=\theta$. That means that $\sin(\theta) = u$, and since $\sin^2(\theta)+\cos^2(\theta)=1$, then $\cos^2(\theta) = 1 - \sin^2(\theta) = 1-u^2$. Therefore, $|\cos(\theta)|=\sqrt{\cos^2\theta} = \sqrt{1-u^2}$; and because in order to talk about the inverse of $\sin \theta$ we must have $-\frac{\pi}{2}\leq \theta\leq \frac{\pi}{2}$, then $\cos\theta\geq 0$, so $|\cos\theta|=\cos\theta$. That is, $\cos\theta = \sqrt{1-u^2}$. So, plugging into the formula for the derivative of $\arcsin(u)$, we have: $\frac{d}{du}\arcsin(u) = \frac{1}{\cos(\arcsin u)} = \frac{1}{\sqrt{1-u^2}}.$
Performing the same kind of analysis for $\arccos(u)$, we get $\frac{d}{du}\arccos(u) = \frac{1}{-\sin(\arccos u)} = -\frac{1}{\sqrt{1-u^2}}.$
For $\arctan u$, using the fact that (\tan\theta)' = \sec^2\theta, we have $\frac{d}{du}\arctan u = \frac{1}{\sec^2(\arctan u)}.$ Now, if $\arctan u = \theta$, then $\tan(\theta) = u$. Using the fact that $\tan^2\theta + 1 = \sec^2\theta$, we get that $sec^2(\arctan u) = \sec^2(\theta) = 1 + \tan^2(\theta) = 1+u^2$, so $\frac{d}{du}\arctan u = \frac{1}{\sec^2(\arctan u)} = \frac{1}{1+u^2}.$
For $\mathrm{arccot u}$, the same analysis works, provided you remember that (\cot\theta)' = -\csc^2\theta and that $1 + \cot^2\theta = \csc^2\theta$, so $\frac{d}{du}\mathrm{arccot}(u) = \frac{1}{-\csc^2(\mathrm{arccot}(u))} = -\frac{1}{1+u^2}.$
With $\mathrm{arcsec}u$, we have (sec\theta)' = sec\theta\tan\theta, so $\frac{d}{du}\mathrm{arcsec}(u) = \frac{1}{\sec(\mathrm{arcsec} (u))\tan(\mathrm{arcsec} u)}.$ Here, $\sec(\mathrm{arcsec} (u)) = u$; if $\mathrm{arcsec}(u)=\theta$, then $\sec\theta = u$, and from $\tan^2\theta + 1 = \sec^2\theta$, we get $|\tan\theta| = \sqrt{u^2 - 1}$. You get: $\frac{d}{du}\mathrm{arcsec}(u) = \frac{1}{\sec(\mathrm{arcsec}(u))\tan(\mathrm{arcsec}(u))} = \frac{1}{u\sqrt{u^2-1}}.$
And finally, using the fact that (\csc\theta)' = -\csc\theta\cot\theta, you get $\frac{d}{du}\mathrm{arccsc}(u) = \frac{1}{-\csc(\mathrm{arccsc}(u))\cot(\mathrm{arccsc}(u))} = -\frac{1}{u\sqrt{u^2-1}}.$
Once you have these formulas, the Chain Rule takes care of the rest.
So you have: \begin{align*} \frac{d}{du}\arcsin(u) &= \frac{1}{\sqrt{1-u^2}}, &\qquad \frac{d}{du}\arccos u &= -\frac{1}{\sqrt{1-u^2}},\\ \frac{d}{du}\arctan(u) &=\frac{1}{1+u^2}, &\frac{d}{du}\mathrm{arccot}(u) &= -\frac{1}{1+u^2},\\ \frac{d}{du}\mathrm{arcsec}(u) &=\frac{1}{u\sqrt{u^2-1}}, &\frac{d}{du}\mathrm{arccsc}(u) &= - \frac{1}{u\sqrt{u^2-1}}. \end{align*}
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0Thanks for your help. Really appreciated. – 2011-02-21
I'll walk you through a derivation (how to get there) of the derivative of the arcsine function. The same idea can be applied to the other inverse trigonometric functions. If $y=\arcsin x$, then $\sin y=\sin(\arcsin x)=x$. $\sin y=x$ Take the derivative of both sides with respect to $x$ (remember the chain rule!). $\cos y\cdot\frac{dy}{dx}=1$ Isolate $\frac{dy}{dx}$, which is what we're trying to find. $\frac{dy}{dx}=\frac{1}{\cos y}$ Since we want $\frac{dy}{dx}$ in terms of $x$, substitute for $y$. $\frac{dy}{dx}=\frac{1}{\cos (\arcsin x)}$ $\cos(\arcsin x)=\sqrt{1-x^2}$ (see this answer of mine for a technique for simplifying a trig function of an inverse trig function), so $\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}.$
When you say "derive these functions" do you mean "take the derivative of these functions?" If so, the chain rule is your friend. So for $f(x)=\arctan(x^2)$, f'(x)=\frac{1}{1+(x^2)^2}2x
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1@Pacane: From $\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}$. – 2011-02-21
if $y=\arcsin(x)$ then $\sin(y)=x$ and differentiating wrt $x$ we get cos(y)y'=1. So, y'=\frac{1}{\cos(\arcsin(x))}. draw a triangle (with sides 1, $x$, $\sqrt{1-x^2}$) to see that $\cos(\arcsin(x))=\sqrt{1-x^2}$. Hence $\frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}$. This kind of reasoning works for various inverse function (using implicit differentiation). see any calc textbook under implicit differentiation.