Math Doubts

Derivatives of Trigonometric functions

In differential calculus, there are six derivative formulas to find the differentiation of the trigonometric functions. Each derivative rule is given here with mathematical proof.

Trigonometric formulas

$(1).\,$ $\dfrac{d}{dx}\,\sin{x}$ $\,=\,$ $\cos{x}$

$(2).\,$ $\dfrac{d}{dx}\,\cos{x}$ $\,=\,$ $-\sin{x}$

$(3).\,$ $\dfrac{d}{dx}\,\tan{x}$ $\,=\,$ $\sec^2{x}$

$(4).\,$ $\dfrac{d}{dx}\,\cot{x}$ $\,=\,$ $-\csc^2{x}$ (or) $-\operatorname{cosec}^2{x}$

$(5).\,$ $\dfrac{d}{dx}\,\sec{x}$ $\,=\,$ $\sec{x}.\tan{x}$

$(6).\,$ $\dfrac{d}{dx}\,\csc{x}$ $\,=\,$ $-\csc{x}.\cot{x}$ (or) $-\operatorname{cosec}{x}.\cot{x}$

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Multiple angle Trigonometric rules

$(1).\,$ $\dfrac{d}{dx}{\,\sin{(ax)}}$ $\,=\,$ $a\cos{(ax)}$

$(2).\,$ $\dfrac{d}{dx}{\,\cos{(ax)}} \,=\, -a\sin{(ax)}$

$(3).\,$ $\dfrac{d}{dx}{\,\tan{(ax)}} \,=\, a\sec^2{(ax)}$

$(4).\,$ $\dfrac{d}{dx}{\,\cot{(ax)}}$ $\,=\,$ $-a\csc^2{(ax)}$ (or) $-a\operatorname{cosec}^2{(ax)}$

$(5).\,$ $\dfrac{d}{dx}{\,\sec{(ax)}} \,=\, a\sec{(ax)}.\tan{(ax)}$

$(6).\,$ $\dfrac{d}{dx}{\,\csc{(ax)}}$ $\,=\,$ $-a\csc{(ax)}.\cot{(ax)}$ (or) $-a\operatorname{cosec}{(ax)}.\cot{(ax)}$

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Compound angle Trigonometric laws

$(1).\,$ $\dfrac{d}{dx}{\,\sin{(ax+b)}}$ $\,=\,$ $a\cos{(ax+b)}$

$(2).\,$ $\dfrac{d}{dx}{\,\cos{(ax+b)}} \,=\, -a\sin{(ax+b)}$

$(3).\,$ $\dfrac{d}{dx}{\,\tan{(ax+b)}} \,=\, a\sec^2{(ax+b)}$

$(4).\,$ $\dfrac{d}{dx}{\,\cot{(ax+b)}}$ $\,=\,$ $-a\csc^2{(ax+b)}$ (or) $-a\operatorname{cosec}^2{(ax+b)}$

$(5).\,$ $\dfrac{d}{dx}{\,\sec{(ax+b)}} \,=\, a\sec{(ax+b)}.\tan{(ax+b)}$

$(6).\,$ $\dfrac{d}{dx}{\,\csc{(ax+b)}}$ $\,=\,$ $-a\csc{(ax+b)}.\cot{(ax+b)}$ (or) $-a\operatorname{cosec}{(ax+b)}.\cot{(ax+b)}$

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