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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​

Define inverse trigonometric functions.

Functions that 'undo' the regular trigonometric functions. If sin⁡(y)=x\sin(y) = xsin(y)=x, then y=sin⁡−1(x)y = \sin^{-1}(x)y=sin−1(x).

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

The inverse sine function, which returns the angle whose sine is x.

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

The inverse cosine function, which returns the angle whose cosine is x.

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

The inverse tangent function, which returns the angle whose tangent is x.

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

The inverse cosecant function, which returns the angle whose cosecant is x.

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

The inverse secant function, which returns the angle whose secant is x.

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

The inverse cotangent function, which returns the angle whose cotangent is x.

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​