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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 you differentiate sin⁡−1(u(x))\sin^{-1}(u(x))sin−1(u(x)) where u(x) is a function of x?

  1. Apply the chain rule: ddx[sin⁡−1(u(x))]=11−(u(x))2⋅u′(x)\frac{d}{dx} [\sin^{-1}(u(x))] = \frac{1}{\sqrt{1 - (u(x))^2}} \cdot u'(x)dxd​[sin−1(u(x))]=1−(u(x))2​1​⋅u′(x). 2. Simplify the expression.

How do you differentiate tan⁡−1(u(x))\tan^{-1}(u(x))tan−1(u(x)) where u(x) is a function of x?

  1. Apply the chain rule: ddx[tan⁡−1(u(x))]=11+(u(x))2⋅u′(x)\frac{d}{dx} [\tan^{-1}(u(x))] = \frac{1}{1 + (u(x))^2} \cdot u'(x)dxd​[tan−1(u(x))]=1+(u(x))21​⋅u′(x). 2. Simplify the expression.

Steps to find the derivative of cos⁡−1(2x)\cos^{-1}(2x)cos−1(2x)?

  1. Apply the chain rule: ddx[cos⁡−1(2x)]=−11−(2x)2⋅2\frac{d}{dx} [\cos^{-1}(2x)] = -\frac{1}{\sqrt{1 - (2x)^2}} \cdot 2dxd​[cos−1(2x)]=−1−(2x)2​1​⋅2. 2. Simplify: −21−4x2\frac{-2}{\sqrt{1-4x^2}}1−4x2​−2​

Steps to find the derivative of tan⁡−1(x2)\tan^{-1}(x^2)tan−1(x2)?

  1. Apply the chain rule: ddx[tan⁡−1(x2)]=11+(x2)2⋅2x\frac{d}{dx} [\tan^{-1}(x^2)] = \frac{1}{1 + (x^2)^2} \cdot 2xdxd​[tan−1(x2)]=1+(x2)21​⋅2x. 2. Simplify: 2x1+x4\frac{2x}{1+x^4}1+x42x​

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​