Inverse Trig Functions

Outer layer $\arctan$, inside $u = x^3$.

$\dfrac{1}{1 + (x^3)^2} = \dfrac{1}{1 + x^6}$.

$u' = 3x^2$.

$f'(x) = \dfrac{3x^2}{1 + x^6}$.

A product of $x$ and $\arcsin u$ with $u = 2x$.

$\dfrac{1}{\sqrt{1 - (2x)^2}} = \dfrac{1}{\sqrt{1 - 4x^2}}$.

$u' = 2$, so $\big(\arcsin(2x)\big)' = \dfrac{2}{\sqrt{1 - 4x^2}}$.

Product rule: $g'(x) = \arcsin(2x) + \dfrac{2x}{\sqrt{1 - 4x^2}}$.

Outer layer $\arccos$, inside $u = \sqrt{x}$.

$-\dfrac{1}{\sqrt{1 - (\sqrt{x})^2}} = -\dfrac{1}{\sqrt{1 - x}}$.

$u' = \dfrac{1}{2\sqrt{x}}$.

$h'(x) = -\dfrac{1}{2\sqrt{x}\sqrt{1 - x}}$.

$\theta = \arctan x$, so $\tan\theta = x$.

$\tan\theta = \dfrac{x}{1}$: opposite $x$, adjacent $1$.

Hypotenuse $\sqrt{1 + x^2}$.

$f(x) = \sin\theta = \dfrac{x}{\sqrt{1 + x^2}}$.

Quotient rule: $f'(x) = \dfrac{\sqrt{1 + x^2} - \frac{x^2}{\sqrt{1 + x^2}}}{1 + x^2} = \dfrac{1}{(1 + x^2)^{3/2}}$. So $f(1) = \dfrac{1}{\sqrt{2}}$, $f'(1) = \dfrac{1}{2\sqrt{2}}$, and $y - \dfrac{1}{\sqrt{2}} = \dfrac{1}{2\sqrt{2}}(x - 1)$.

$\theta = \operatorname{arcsec} x$, so $\sec\theta = x$, that is $\cos\theta = \dfrac{1}{x}$.

Adjacent $1$, hypotenuse $x$.

Opposite $\sqrt{x^2 - 1}$.

$g(x) = \tan\theta = \dfrac{\sqrt{x^2 - 1}}{1} = \sqrt{x^2 - 1}$ (positive, since $0 < \theta < \tfrac{\pi}{2}$ for $x > 1$).

$g'(x) = \dfrac{2x}{2\sqrt{x^2 - 1}} = \dfrac{x}{\sqrt{x^2 - 1}}$.

$\theta = \arctan 2x$, so $\tan\theta = 2x$.

$\tan\theta = \dfrac{2x}{1}$: opposite $2x$, adjacent $1$.

Hypotenuse $\sqrt{1 + 4x^2}$.

$h(x) = \cos\theta = \dfrac{1}{\sqrt{1 + 4x^2}} = (1 + 4x^2)^{-1/2}$.

$h'(x) = -\tfrac{1}{2}(1 + 4x^2)^{-3/2} \cdot 8x = -\dfrac{4x}{(1 + 4x^2)^{3/2}}$. So $h(1) = \dfrac{1}{\sqrt{5}}$, $h'(1) = -\dfrac{4}{5\sqrt{5}}$, and $y - \dfrac{1}{\sqrt{5}} = -\dfrac{4}{5\sqrt{5}}(x - 1)$.

$y = \arcsin x$ means $\sin y = x$, with $-\tfrac{\pi}{2} \leq y \leq \tfrac{\pi}{2}$.

$\cos y \cdot y' = 1$.

$y' = \dfrac{1}{\cos y}$.

$\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ (positive, since $\cos y \geq 0$ for $y$ in $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$). So $y' = \dfrac{1}{\sqrt{1 - x^2}}$.

$y = f^{-1}(x)$ means $y^3 + 2y + 1 = x$.

$(3y^2 + 2)\,y' = 1$.

$y' = \dfrac{1}{3y^2 + 2}$.

At $x = 4$: solve $y^3 + 2y + 1 = 4$; try $y = 1$: $1 + 2 + 1 = 4$. So $(f^{-1})'(4) = \dfrac{1}{3 + 2} = \dfrac{1}{5}$.

$y = \arctan x$ means $\tan y = x$, with $-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}$.

$\sec^2 y \cdot y' = 1$.

$y' = \dfrac{1}{\sec^2 y}$.

$\sec^2 y = 1 + \tan^2 y = 1 + x^2$, so $y' = \dfrac{1}{1 + x^2}$.