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How do the derivatives of sin⁡−1(x)\sin^{-1}(x) and cos⁡−1(x)\cos^{-1}(x) relate?

ddx[cos⁡−1(x)]\frac{d}{dx}[\cos^{-1}(x)] is the negative of ddx[sin⁡−1(x)]\frac{d}{dx}[\sin^{-1}(x)]. Specifically, ddx[sin⁡−1(x)]=11−x2\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}} and ddx[cos⁡−1(x)]=−11−x2\frac{d}{dx}[\cos^{-1}(x)] = -\frac{1}{\sqrt{1-x^2}}

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How do the derivatives of $\sin^{-1}(x)$ and $\cos^{-1}(x)$ relate?
$\frac{d}{dx}[\cos^{-1}(x)]$ is the negative of $\frac{d}{dx}[\sin^{-1}(x)]$. Specifically, $\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}}$ and $\frac{d}{dx}[\cos^{-1}(x)] = -\frac{1}{\sqrt{1-x^2}}$
How do the derivatives of $\tan^{-1}(x)$ and $\cot^{-1}(x)$ relate?
$\frac{d}{dx}[\cot^{-1}(x)]$ is the negative of $\frac{d}{dx}[\tan^{-1}(x)]$. Specifically, $\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2}$ and $\frac{d}{dx}[\cot^{-1}(x)] = -\frac{1}{1+x^2}$
How do the derivatives of $\sec^{-1}(x)$ and $\csc^{-1}(x)$ relate?
$\frac{d}{dx}[\csc^{-1}(x)]$ is the negative of $\frac{d}{dx}[\sec^{-1}(x)]$. Specifically, $\frac{d}{dx}[\sec^{-1}(x)] = \frac{1}{\lvert{x}\rvert\sqrt{x^{2}-1}}$ and $\frac{d}{dx}[\csc^{-1}(x)] = -\frac{1}{\lvert{x}\rvert\sqrt{x^{2}-1}}$
How do you differentiate $\sin^{-1}(u(x))$ where u(x) is a function of x?
1. Apply the chain rule: $\frac{d}{dx} [\sin^{-1}(u(x))] = \frac{1}{\sqrt{1 - (u(x))^2}} \cdot u'(x)$. 2. Simplify the expression.
How do you differentiate $\tan^{-1}(u(x))$ where u(x) is a function of x?
1. Apply the chain rule: $\frac{d}{dx} [\tan^{-1}(u(x))] = \frac{1}{1 + (u(x))^2} \cdot u'(x)$. 2. Simplify the expression.
Steps to find the derivative of $\cos^{-1}(2x)$?
1. Apply the chain rule: $\frac{d}{dx} [\cos^{-1}(2x)] = -\frac{1}{\sqrt{1 - (2x)^2}} \cdot 2$. 2. Simplify: $\frac{-2}{\sqrt{1-4x^2}}$
Steps to find the derivative of $\tan^{-1}(x^2)$?
1. Apply the chain rule: $\frac{d}{dx} [\tan^{-1}(x^2)] = \frac{1}{1 + (x^2)^2} \cdot 2x$. 2. Simplify: $\frac{2x}{1+x^4}$
What is the derivative of $\sin^{-1}(x)$?
$\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}}$
What is the derivative of $\cos^{-1}(x)$?
$\frac{d}{dx}[\cos^{-1}(x)] = -\frac{1}{\sqrt{1-x^2}}$
What is the derivative of $\tan^{-1}(x)$?
$\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2}$
What is the derivative of $\csc^{-1}(x)$?
$\frac{d}{dx}[\csc^{-1}(x)] = -\frac{1}{\lvert{x}\rvert\sqrt{x^{2}-1}}$
What is the derivative of $\sec^{-1}(x)$?
$\frac{d}{dx}[\sec^{-1}(x)] = \frac{1}{\lvert{x}\rvert\sqrt{x^{2}-1}}$
What is the derivative of $\cot^{-1}(x)$?
$\frac{d}{dx}[\cot^{-1}(x)] = -\frac{1}{1+x^2}$
State the general formula for the derivative of an inverse function.
$\frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))}$