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Revision:Differentiation of Trigonometry Functions

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Differentiation of trig functions

Contents

What you need to know

It is possible to find the derivative of trigonometric functions. Here is a list of the standard forms that you need to know:

\displaystyle \frac{d (\sin x)}{dx} = \cos x


\displaystyle \frac{d (\cos x)}{dx} = -\sin x


\displaystyle \frac{d (\sec x)}{dx} = \sec x \tan x


\displaystyle \frac{d (\csc x)}{dx} = -\csc x \cot x


\displaystyle \frac{d (\tan x)}{dx} = \sec ^2 x


\displaystyle \frac{d (\cot x)}{dx} = -\csc ^2 x


One condition upon these results is that x must be measured in radians.

Proof

To differentiate sin x from first principles, let us consider the gradient m of the line between the points (x, sin x) and (x+o, sin(x+o)), where o is small.

\displaystyle m = \frac{\sin (x+o) - \sin x}{x+o - x} = \frac{\sin (x+o) - \sin x}{o}

\displaystyle = \frac{2\cos \tfrac{2x+o}{2} \sin \tfrac{o}{2}}{o} = \frac{\cos \tfrac{2x+o}{2} \sin \tfrac{o}{2}}{\tfrac{o}{2}}

\displaystyle =\cos (x + \tfrac{o}{2} ) \frac{\sin (o/2)}{o/2}

As o tends to 0, [sin(o/2)]/(o/2) tends to 1, and cos(x + o/2) tends to cos x, and so m tends to cos x; that is,

\dfrac{\text{d}}{\text{d} x} \sin x = \cos x.

Remember that cos x = sin(π/2 - x). Replacing x with (π/2 - x), and using the chain rule, in the above:

\dfrac{\text{d}}{\text{d} x} \sin ({\pi\over 2} - x) = -\cos ({\pi\over 2} - x).

That is,

\dfrac{\text{d}}{\text{d} x} \cos x = -\sin x.

Using the quotient rule:

\displaystyle \frac{\text{d}}{\text{d} x} \tan x = \frac{\text{d}}{\text{d} x} \frac{\sin x}{\cos x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \sec^2 x.

\displaystyle \frac{\text{d}}{\text{d} x} \sec x = \frac{\text{d}}{\text{d} x} \frac{1}{\cos x} = \frac{\sin x}{\cos^2 x} = \sec x\tan x.

Deriving the final two by noting that cot x = tan(π/2 - x) and csc x = sec(π/2 - x) respectively:

\displaystyle \frac{\text{d}}{\text{d} x} \cot x = -\csc^2 x, \;\;\;\; \frac{\text{d}}{\text{d} x} \csc x = -\csc x\cot x.


Applying the Chain Rule: An example

The chain rule is used to differentiate harder trigonometric functions.


Differentiate \cos ^3 x with respect to x.

Let \displaystyle y = \cos ^3 x


Let \displaystyle u = \cos x

therefore \displaystyle y = u^3 \displaystyle \frac{dy}{du} = 3u^2

\displaystyle \frac{du}{dx} = -\sin x


\displaystyle \frac{dy}{dx} = \frac{du}{dx} \times \frac{dy}{du}

\displaystyle = -\sin x \times 3u^2

\displaystyle = -\sin x \times 3\cos ^2 x

\displaystyle = -3\cos ^2 x \sin x


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