disclamer: I won't be proving any of those cuz it is such a brain-melting pain to prove those and write the latex code of the proof, however, I proved \(\frac{d}{dx} \sin(x)\) and \(\frac{d}{dx} \cos(x)\) only. Additionally, I won't be able to show all trig derivatives because of the sheer amount of them. Like why the hell should I know what is \(\frac{d}{dx} \tanh(x)\) (btw it is \(\operatorname{sech}^2(x)\)) If u fade something like this in an exam, u should try to find a simplification using normal trig identities, or just plug it into the general formula: \(\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\)
Ok, I HATE trig identities. If u r a functioning human being, you propably hate trig identities, so I will rapid fire through trig differentiation $$\frac{d}{dx} \sin(x) = \cos(x)$$ $$\frac{d}{dx} \cos(x) = -\sin(x)$$ $$\frac{d}{dx} \tan(x) = \sec^2(x)$$ $$\frac{d}{dx} \cot(x) = -\csc^2(x)$$ $$\frac{d}{dx} \sec(x) = \sec(x) \tan(x)$$ $$\frac{d}{dx} \csc(x) = -\csc(x) \cot(x)$$ $$\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}$$ $$\frac{d}{dx} \cos^{-1}(x) = -\frac{1}{\sqrt{1-x^2}}$$ $$\frac{d}{dx} \tan^{-1}(x) = \frac{1}{1+x^2}$$ $$\frac{d}{dx} \cot^{-1}(x) = -\frac{1}{1+x^2}$$ And now for the 2 most cursed Identities of all: $$\frac{d}{dx} \sec^{-1}(x) = \frac{1}{|x|\sqrt{x^2-1}}$$ $$\frac{d}{dx} \csc^{-1}(x) = -\frac{1}{|x|\sqrt{x^2-1}}$$