$$\begin{aligned} \dfrac{d}{dx} sin(x) &= cos(x) \\ \\ \dfrac{d}{dx} tan(x) &= sec^2(x) \\ \\ \dfrac{d}{dx} sec(x) &= sec(x) \cdot tan(x) \\ \end{aligned}$$

$$\begin{aligned} \dfrac{d}{dx} cos(x) &= -sin(x) \\ \\ \dfrac{d}{dx} cot(x) &= -csc^2(x) \\ \\ \dfrac{d}{dx} csc(x) &= -csc(x) \cdot cot(x) \\ \end{aligned}$$

$$\begin{aligned} \dfrac{d}{dx} sin^{-1}(x) &= \frac{1}{\sqrt{1 - x^2}} \\ \\ \dfrac{d}{dx} tan^{-1}(x) &= \frac{1}{1 + x^2} \\ \\ \dfrac{d}{dx} sec^{-1}(x) &= \frac{1}{x \cdot \sqrt{x^2 - 1}} \\ \end{aligned}$$

$$\begin{aligned} \dfrac{d}{dx} cos^{-1}(x) &= \frac{-1}{\sqrt{1 - x^2}} \\ \\ \dfrac{d}{dx} cot^{-1}(x) &= \frac{-1}{1 + x^2} \\ \\ \dfrac{d}{dx} csc^{-1}(x) &= \frac{-1}{x \cdot \sqrt{x^2 - 1}} \\ \end{aligned}$$

Take the last trig function as an example. Let $f(x) = csc^{-1}(x)$,

$$\begin{aligned} f(csc(x)) &= csc^{-1}(csc(x)) = x \\ f'(csc(x)) \cdot csc'(x) &= 1 \\ f'(csc(x)) &= \frac{1}{csc'(x)} \\ &= \frac{-1}{csc(x) \cdot cot(x)} \\ \end{aligned}$$

Since $sin^2(x) + cos^2(x) = 1$, divide both sides by $sin^2(x)$:

$$\begin{aligned} 1 + cot^2(x) &= csc^2(x) \\ cot(x) &= \sqrt{csc^2(x) - 1}\\ \end{aligned}$$

From above,

$$\begin{aligned} f'(csc(x)) &= \frac{-1}{csc(x) \cdot cot(x)} \\ f'(csc(x)) &= \frac{-1}{csc(x) \cdot \sqrt{csc^2(x) - 1}} \\ f'(x) &= \frac{-1}{x \cdot \sqrt{x^2 - 1}} \\ \therefore \dfrac{d}{dx} csc^{-1}(x) &= \frac{-1}{x \cdot \sqrt{x^2 - 1}} \\ \end{aligned}$$

# posted by rot13(Unafba Pune) @ 10:48 PM