• Math Doubts
• Differentiation
• Formulae

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) $-\operatorname{cosec}{x}.\cot{x}$

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) $-a\operatorname{cosec}{(ax)}.\cot{(ax)}$

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) $-a\operatorname{cosec}{(ax+b)}.\cot{(ax+b)}$

Ashok Kumar, B.E.

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