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.
[Math] Trying to derive an inverse trigonometric function
calculustrigonometry
Related Solutions
I'll be ignoring domains and possible roots of negative numbers. (If you let $\mbox{$x\in \textbf{]}0,\pi/2[$}$ everything works fine).
Given $f\circ g$, the trick is too relate $f$ with $g^{-1}$.
I did some. You should be able to handle the remaining ones.
$\bullet \sin (\arccos (x))=\sqrt {1-(\cos (\arccos (x) ))^2}=\sqrt {1-x^2}$
$\bullet \sin (\arctan (x))=\dfrac{\tan (\arctan (x))}{\sqrt {1+(\tan (\arctan (x)))^2}}=\dfrac{x}{\sqrt {1+x^2}}$
For this one I used $$\begin{align}(\cos(x))^2+(\sin (x))^2=1 &\implies 1+(\tan(x))^2=(\sec(x))^2\\ &\implies 1+(\tan (x))^2=\dfrac{1}{1-(\sin (x))^2}\\ &\implies 1-(\sin(x))^2=\dfrac{1}{1+(\tan(x))^2}\\ &\implies \sin (x)=\dfrac{\tan(x)}{\sqrt{1+(\tan(x))^2}}\end{align}$$
$\bullet \tan (\arcsin (x))=\dfrac{\sin (\arcsin (x))}{\sqrt{1-(\sin(\arcsin (x)))^2}}=\dfrac{x}{\sqrt{1-x^2}}$
For this one I used $$\begin{align}(\cos(x))^2+(\sin (x))^2=1 &\implies 1+(\tan(x))^2=(\sec(x))^2\\ &\implies 1+(\tan (x))^2=\dfrac{1}{1-(\sin (x))^2}\\ &\implies \tan(x)=\dfrac{\sin(x)}{\sqrt{1-(\sin(x))^2}}\end{align}$$
$\bullet \cot (\arcsin(x))=\dfrac{\sqrt{1-(\sin (\arcsin(x)))^2}}{\sin (\arcsin(x))}=\dfrac{\sqrt{1-x^2}}{x}$
For this one I used $$\begin{align}(\cos(x))^2+(\sin (x))^2=1 &\implies (\cot (x))^2+1=(\csc(x))^2\\ &\implies (\cot (x))^2=\dfrac{1-(\sin(x))^2}{(\sin(x))^2}\\ &\implies \cot(x)=\dfrac{\sqrt{1-(\sin(x))^2}}{\sin (x)}\end{align}$$
$\bullet \cot (\arccos(x))=\dfrac{\cos (\arccos(x))}{\sqrt{1-(\cos (\arccos(x)))^2}}=\dfrac{x}{\sqrt{1-x^2}}$.
For this one I used $$\begin{align}(\cos(x))^2+(\sin (x))^2=1 &\implies (\cot (x))^2+1=(\csc(x))^2\\ &\implies (\cot (x))^2=\dfrac{1-(\sin(x))^2}{(\sin(x))^2}\\ &\implies (\cot(x))^2=\dfrac{(\cos(x))^2}{1-(\cos(x))^2}\\ &\implies \cot(x)=\dfrac{\cos(x)}{\sqrt{1-(\cos(x))^2}}\end{align}$$
You might find these identities useful
Related Question
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Best Answer
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*}