Trying to derive an inverse trigonometric function 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.
 A: 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 $$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}.$$
A: 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 $f(x)=\arctan(x^2)$, $f'(x)=\frac{1}{1+(x^2)^2}2x$
A: 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*}
A: 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.
