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Calculus (Guichard)
4: Transcendental Functions
4.5: Derivatives of the Trigonometric Functions

4.5: Derivatives of the Trigonometric Functions

Last updated

Feb 8, 2025
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- 4.4: The Derivative of sin x - II
- 4.6: Exponential and Logarithmic Functions

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All of the other trigonometric functions can be expressed in terms of the sine, and so their derivatives can easily be calculated using the rules we already have. For the cosine we need to use two identities,

$\eqalign{ \cos x &= \sin(x+{\pi\over2}),\cr \sin x &= -\cos(x+{\pi\over2}).\cr }$

Now:

$\eqalign{ {d\over dx}\cos x &= {d\over dx}\sin \left(x+{\pi\over2}\right) = \cos \left(x+{\pi\over2}\right )\cdot 1 = -\sin x\cr {d\over dx}\tan x &= {d\over dx}{\sin x\over \cos x}= {\cos^2 x + \sin^2 x\over \cos^2 x}={1\over \cos^2 x}=\sec^2 x\cr {d\over dx}\sec x &= {d\over dx}(\cos x)^{-1}= -1(\cos x)^{-2}(-\sin x) = {\sin x \over \cos^2 x} = \sec x\tan x.\cr }$

The derivatives of the cotangent and cosecant are similar and left as exercises.

Contributors

David Guichard (Whitman College)
Integrated by Justin Marshall.

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