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How do the derivatives of sin1(x)\sin^{-1}(x) and cos1(x)\cos^{-1}(x) relate?

ddx[cos1(x)]\frac{d}{dx}[\cos^{-1}(x)] is the negative of ddx[sin1(x)]\frac{d}{dx}[\sin^{-1}(x)]. Specifically, ddx[sin1(x)]=11x2\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}} and ddx[cos1(x)]=11x2\frac{d}{dx}[\cos^{-1}(x)] = -\frac{1}{\sqrt{1-x^2}}

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How do the derivatives of sin1(x)\sin^{-1}(x) and cos1(x)\cos^{-1}(x) relate?

ddx[cos1(x)]\frac{d}{dx}[\cos^{-1}(x)] is the negative of ddx[sin1(x)]\frac{d}{dx}[\sin^{-1}(x)]. Specifically, ddx[sin1(x)]=11x2\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}} and ddx[cos1(x)]=11x2\frac{d}{dx}[\cos^{-1}(x)] = -\frac{1}{\sqrt{1-x^2}}

How do the derivatives of tan1(x)\tan^{-1}(x) and cot1(x)\cot^{-1}(x) relate?

ddx[cot1(x)]\frac{d}{dx}[\cot^{-1}(x)] is the negative of ddx[tan1(x)]\frac{d}{dx}[\tan^{-1}(x)]. Specifically, ddx[tan1(x)]=11+x2\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2} and ddx[cot1(x)]=11+x2\frac{d}{dx}[\cot^{-1}(x)] = -\frac{1}{1+x^2}

How do the derivatives of sec1(x)\sec^{-1}(x) and csc1(x)\csc^{-1}(x) relate?

ddx[csc1(x)]\frac{d}{dx}[\csc^{-1}(x)] is the negative of ddx[sec1(x)]\frac{d}{dx}[\sec^{-1}(x)]. Specifically, ddx[sec1(x)]=1xx21\frac{d}{dx}[\sec^{-1}(x)] = \frac{1}{\lvert{x}\rvert\sqrt{x^{2}-1}} and ddx[csc1(x)]=1xx21\frac{d}{dx}[\csc^{-1}(x)] = -\frac{1}{\lvert{x}\rvert\sqrt{x^{2}-1}}

What is the derivative of sin1(x)\sin^{-1}(x)?

ddx[sin1(x)]=11x2\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}}

What is the derivative of cos1(x)\cos^{-1}(x)?

ddx[cos1(x)]=11x2\frac{d}{dx}[\cos^{-1}(x)] = -\frac{1}{\sqrt{1-x^2}}

What is the derivative of tan1(x)\tan^{-1}(x)?

ddx[tan1(x)]=11+x2\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2}

What is the derivative of csc1(x)\csc^{-1}(x)?

ddx[csc1(x)]=1xx21\frac{d}{dx}[\csc^{-1}(x)] = -\frac{1}{\lvert{x}\rvert\sqrt{x^{2}-1}}

What is the derivative of sec1(x)\sec^{-1}(x)?

ddx[sec1(x)]=1xx21\frac{d}{dx}[\sec^{-1}(x)] = \frac{1}{\lvert{x}\rvert\sqrt{x^{2}-1}}

What is the derivative of cot1(x)\cot^{-1}(x)?

ddx[cot1(x)]=11+x2\frac{d}{dx}[\cot^{-1}(x)] = -\frac{1}{1+x^2}

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

ddx[f1(x)]=1f(f1(x))\frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))}

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 sin1(x)\sin^{-1}(x).

Start with y=sin1(x)y = \sin^{-1}(x), rewrite as x=sin(y)x = \sin(y), differentiate both sides with respect to x, and solve for dydx\frac{dy}{dx}.