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  2. AP Calculus
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What is the derivative of sin⁡−1(x)\sin^{-1}(x)sin−1(x)?

ddx[sin⁡−1(x)]=11−x2\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}}dxd​[sin−1(x)]=1−x2​1​

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What is the derivative of sin⁡−1(x)\sin^{-1}(x)sin−1(x)?

ddx[sin⁡−1(x)]=11−x2\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}}dxd​[sin−1(x)]=1−x2​1​

What is the derivative of cos⁡−1(x)\cos^{-1}(x)cos−1(x)?

ddx[cos⁡−1(x)]=−11−x2\frac{d}{dx}[\cos^{-1}(x)] = -\frac{1}{\sqrt{1-x^2}}dxd​[cos−1(x)]=−1−x2​1​

What is the derivative of tan⁡−1(x)\tan^{-1}(x)tan−1(x)?

ddx[tan⁡−1(x)]=11+x2\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2}dxd​[tan−1(x)]=1+x21​

What is the derivative of csc⁡−1(x)\csc^{-1}(x)csc−1(x)?

ddx[csc⁡−1(x)]=−1∣x∣x2−1\frac{d}{dx}[\csc^{-1}(x)] = -\frac{1}{\lvert{x}\rvert\sqrt{x^{2}-1}}dxd​[csc−1(x)]=−∣x∣x2−1​1​

What is the derivative of sec⁡−1(x)\sec^{-1}(x)sec−1(x)?

ddx[sec⁡−1(x)]=1∣x∣x2−1\frac{d}{dx}[\sec^{-1}(x)] = \frac{1}{\lvert{x}\rvert\sqrt{x^{2}-1}}dxd​[sec−1(x)]=∣x∣x2−1​1​

What is the derivative of cot⁡−1(x)\cot^{-1}(x)cot−1(x)?

ddx[cot⁡−1(x)]=−11+x2\frac{d}{dx}[\cot^{-1}(x)] = -\frac{1}{1+x^2}dxd​[cot−1(x)]=−1+x21​

State the general formula for the derivative of an inverse function.

ddx[f−1(x)]=1f′(f−1(x))\frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))}dxd​[f−1(x)]=f′(f−1(x))1​

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

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

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

ddx[cot⁡−1(x)]\frac{d}{dx}[\cot^{-1}(x)]dxd​[cot−1(x)] is the negative of ddx[tan⁡−1(x)]\frac{d}{dx}[\tan^{-1}(x)]dxd​[tan−1(x)]. Specifically, ddx[tan⁡−1(x)]=11+x2\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2}dxd​[tan−1(x)]=1+x21​ and ddx[cot⁡−1(x)]=−11+x2\frac{d}{dx}[\cot^{-1}(x)] = -\frac{1}{1+x^2}dxd​[cot−1(x)]=−1+x21​

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

ddx[csc⁡−1(x)]\frac{d}{dx}[\csc^{-1}(x)]dxd​[csc−1(x)] is the negative of ddx[sec⁡−1(x)]\frac{d}{dx}[\sec^{-1}(x)]dxd​[sec−1(x)]. Specifically, ddx[sec⁡−1(x)]=1∣x∣x2−1\frac{d}{dx}[\sec^{-1}(x)] = \frac{1}{\lvert{x}\rvert\sqrt{x^{2}-1}}dxd​[sec−1(x)]=∣x∣x2−1​1​ and ddx[csc⁡−1(x)]=−1∣x∣x2−1\frac{d}{dx}[\csc^{-1}(x)] = -\frac{1}{\lvert{x}\rvert\sqrt{x^{2}-1}}dxd​[csc−1(x)]=−∣x∣x2−1​1​

Explain the core concept of inverse trig derivatives.

Inverse trig functions 'undo' regular trig functions. We find the derivatives of these inverse functions using implicit differentiation and trigonometric identities.

Why is the chain rule important when differentiating inverse trig functions?

Inverse trig functions are often part of composite functions, so the chain rule is necessary to differentiate the 'inside' function.

Explain how implicit differentiation is used to find the derivative of sin⁡−1(x)\sin^{-1}(x)sin−1(x).

Start with y=sin⁡−1(x)y = \sin^{-1}(x)y=sin−1(x), rewrite as x=sin⁡(y)x = \sin(y)x=sin(y), differentiate both sides with respect to x, and solve for dydx\frac{dy}{dx}dxdy​.