5DDerivInvTrig

toc = Derivatives of Inverse Trig Functions =

Arcsin(x)
Recall that the Inverse Sine function, // y = Sin -1 x // (or //y = arsin( x)// ) has a domain of **[-1, 1]**

You are not expected to be able to obtain the derivatives of inverse trig functions, just use them.

math . \qquad \dfrac {d}{dx} \left( \sin^{-1} x \right) = \dfrac {1}{\sqrt{1-x^2}} \; \text{ for } x \in (-1, 1) math

Notice that the endpoints are excluded for two reasons:
 * Derivatives are never defined for the endpoints of a graph
 * If we substitute x = ±1 into the derivative, we get divide by zero, so the derivative is undefined at those points.

Applying the Chain Rule to this formula, gives us math . \qquad \dfrac{d}{dx} \left( \sin^{-1} g(x) \right) = \dfrac {g'(x)}{\sqrt{1-[g(x)]^2}} \; \text{ for } g(x) \in (-1, 1) math {Notice g(x) in the limits}

Example

math . \qquad \dfrac{d}{dx} \left( \sin^{-1} e^{2x} \right) = \dfrac {2e^{2x}}{\sqrt{1-[e^{2x}]^2}} \; \text{ for } e^{2x} \in (-1, 1) math . math . \qquad \dfrac{d}{dx} \left( \sin^{-1} e^{2x} \right) = \dfrac {2e^{2x}}{\sqrt{1-e^{4x}}} \; \text{ for } x \in (-\infty, 0) math

Observe the change in limits from e 2x to x math . \qquad e^{2x} > 0 \text { so this changes limits to: } e^{2x} \in (0, 1) math

Take the natural log (**log e ** or **ln**) of both sides, makes the limits: math . \qquad 2x \in (-\infty, 0) math

Divide by 2, the limits become: math . \qquad x \in (-\infty, 0) math

A different version of the derivative of Arsin We can use the chain rule version to get a new rule:

math . \qquad \dfrac{d}{dx} \left( \sin^{-1} g(x) \right) = \dfrac {g'(x)}{\sqrt{1-[g(x)]^2}} \; \text{ for } g(x) \in (-1, 1) math . math . \qquad \dfrac{d}{dx} \left( \sin^{-1} \dfrac {x}{a} \right) = \dfrac {\dfrac{1}{a}}{\sqrt{1- \left[ \dfrac{x}{a} \right] ^2}} \; \text{ for } \dfrac{x}{a} \in (-1, 1) math . math . \qquad \qquad \qquad \qquad = \dfrac {1}{a\sqrt{1-\dfrac{x^2}{a^2}}} \; \text{ for } x \in (-a, a) math . math . \qquad \qquad \qquad \qquad = \dfrac {1}{\sqrt{a^2 \Big( 1-\dfrac{x^2}{a^2}} \Big) } \; \text{ for } x \in (-a, a) math . math . \qquad \qquad \qquad \qquad = \dfrac {1}{\sqrt{a^2 -x^2}} \; \text{ for } x \in (-a, a) math

So we get the rule: math . \qquad \dfrac{d}{dx} \left( \sin^{-1} \dfrac {x}{a} \right) = \dfrac {1}{\sqrt{a^2 -x^2}} \; \text{ for } x \in (-a, a) math

Example math . \qquad \dfrac{d}{dx} \left( \sin^{-1} 3x \right) = \dfrac {1}{\sqrt{\dfrac{1}{9} -x^2}} \; \text{ for } x \in (-\dfrac{1}{3}, \dfrac{1}{3}) \quad \text {Using } a = \dfrac {1}{3} math

Arccos(x)
In a similar way: math . \qquad \dfrac {d}{dx} \left( \cos^{-1} x \right) = \dfrac {-1}{\sqrt{1-x^2}} \; \text{ for } x \in (-1, 1) math And math . \qquad \dfrac{d}{dx} \left( \cos^{-1} g(x) \right) = \dfrac {-g'(x)}{\sqrt{1-[g(x)]^2}} \; \text{ for } g(x) \in (-1, 1) math And math . \qquad \dfrac{d}{dx} \left( \cos^{-1} \dfrac {x}{a} \right) = \dfrac {-1}{\sqrt{a^2 -x^2}} \; \text{ for } x \in (-a, a) math

Arctan(x)
In a similar way: math . \qquad \dfrac {d}{dx} \left( \tan^{-1} x \right) = \dfrac {1}{1+x^2} \; \text{ for } x \in R math And math . \qquad \dfrac{d}{dx} \left( \tan^{-1} g(x) \right) = \dfrac {g'(x)}{1+[g(x)]^2} \; \text{ for } g(x) \in R math And math . \qquad \dfrac{d}{dx} \left( \tan^{-1} \dfrac {x}{a} \right) = \dfrac {a}{a^2 +x^2} \; \text{ for } x \in R math

These rules can be reversed to obtain rules for antidifferentiation.

**See also** [|www.mathsonline.com.au] Y12Extension --> Functions --> Inverse Functions --> Lesson 4 .