Differentiation of trigonometric functions

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Function	Derivative
$sin(x)$	$cos(x)$
$cos(x)$	$- sin(x)$
$tan(x)$	$sec 2 (x)$
$cot(x)$	$- csc 2 (x)$
$sec(x)$	$sec(x)tan(x)$
$csc(x)$	$- csc(x)cot(x)$
$arcsin(x)$	$\frac{1}{\sqrt{1-x^2}}$
$arccos(x)$	$\frac{-1}{\sqrt{1-x^2}}$
$arctan(x)$	$\frac{1}{x^2+1}$

The differentiation of trigonometric functions is the mathematical process of finding the rate at which a trigonometric function changes with respect to a variable; the derivative of the trigonometric function. Commonplace trigonometric functions include sin(x), cos(x) and tan(x). For example, in differentiating f(x) = sin(x), one is calculating a function f ′(x) which computes the rate of change of sin(x) at a particular point a. The value of the rate of change at a is thus given by f ′(a). Knowledge of differentiation from first principles is required, along with compentence in the use of trigonometric identities and limits. All functions involve the arbitrary variable x, with all differentiation performed with respect to x.

It turns out that once you know the deriatives of sin(x) and cos(x), you can easily compute the derivatives of the other circular trigonometric functions because they can all be expressed in terms of sine or cosine; the quotient rule is then implemented to differentiate this expression. Proofs of the derivatives of sin(x) and cos(x) are given in the proofs section; the results are quoted in order to give proofs of the derivatives of the other circular trigonometric functions. Finding the derivatives of the inverse trigonometric functions involves using implicit differentiation and the derivatives of regular trigonometric functions also given in the proofs section.

1 Derivatives of sin(x), cos(x), tan(x), cot(x), sec(x) and csc(x) and their inverses
2 Proofs of derivative of the sine and cosine functions
- 2.1 Differentiating the sine function
- 2.2 Differentiating the cosine function
3 Proofs of derivatives of inverse trigonometric functions
4 See also

[edit] Derivatives of sin(x), cos(x), tan(x), cot(x), sec(x) and csc(x) and their inverses

$f(x)=\sin(x) \implies f'(x)=\cos(x)$

$f(x)=\cos(x) \implies f'(x)=-\sin(x)$

$f(x)=\tan(x) \implies f'(x) = (\tan(x))' = \left(\frac{\sin(x)}{\cos(x)}\right)' = \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)} = 1+\tan^2(x) = \sec^2(x)$

$f(x)=\cot(x) \implies f'(x) = (\cot(x))' = \left(\frac{\cos(x)}{\sin(x)}\right)' = \frac{-\sin^2(x) - \cos^2(x)}{\sin^2(x)} = -(1+\cot^2(x)) = -\csc^2(x)$

$f(x)=\sec(x) \implies f'(x) = (\sec(x))' = \left(\frac{1}{\cos(x)}\right)' = \frac{\sin(x)}{\cos^2(x)} = \frac{1}{\cos(x)}.\frac{\sin(x)}{\cos(x)} = \sec(x)\tan(x)$

$f(x)=\csc(x) \implies f'(x) = (\csc(x))' = \left(\frac{1}{\sin(x)}\right)' = -\frac{\cos(x)}{\sin^2(x)} = -\frac{1}{\sin(x)}.\frac{\cos(x)}{\sin(x)} = -\csc(x)\cot(x)$

$f(x)=\arcsin(x) \implies f'(x)=\frac{1}{\sqrt{1-x^2}}$

$f(x)=\arccos(x) \implies f'(x)=\frac{-1}{\sqrt{1-x^2}}$

$f(x)=\arctan(x) \implies f'(x)=\frac{1}{x^2+1}$

[edit] Proofs of derivative of the sine and cosine functions

[edit] Differentiating the sine function

Definition of a derivative of a function f(x):

$f'(x)=\lim_{h\to 0}{f(x+h)-f(x)\over h}$

Therefore if f(x) = sin(x)

$f'(x)=\lim_{h\to 0}{\sin(x+h)-\sin(x)\over h}$

From the trigonometric identity

sin(A + B) = sin(A)cos(B) + cos(A)sin(B),

we can say

$f'(x)=\lim_{h\to 0}{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)\over h}$

Collecting together the terms of cos(x) and sin(x), the derivative becomes

$f'(x)=\lim_{h\to 0}{\cos(x)\sin(h)-\sin(x)(1-\cos(h))\over h}$

Rearranging the terms and the limit gives

$f'(x)=\lim_{h\to 0}{\cos(x)\sin(h)\over h} - \lim_{h\to 0}{\sin(x)(1-\cos(h))\over h}$

Now because sin(x) and cos(x) do not vary as h varies, they can be taken out of the limit to give

$f'(x)=\cos(x)\lim_{h\to 0}{\sin(h)\over h} - \sin(x)\lim_{h\to 0}{1-\cos(h)\over h}$

The value of the limits

$\lim_{h\to 0}{\sin(h)\over h} \quad\text{and}\quad \lim_{h\to 0}{1-\cos(h)\over h}$

are 1 and 0 respectively. Therefore, if f(x) = sin(x),

$f'(x)=\cos(x). \,$

[edit] Differentiating the cosine function

Definition of a derivative of a function f(x):

$f'(x)=\lim_{h\to 0}{f(x+h)-f(x)\over h}$

Therefore if f(x) = cos(x)

$f'(x)=\lim_{h\to 0}{\cos(x+h)-\cos(x)\over h}$

From the trigonometric identity

cos(A + B) = cos(A)cos(B) - sin(A)sin(B),

we can say

$f'(x)=\lim_{h\to 0}{\cos(x)\cos(h)-\sin(x)\sin(h)-\cos(x)\over h}$

Collecting together the terms of sin(x) and cos(x), the derivative becomes

$f'(x)=\lim_{h\to 0}{-\sin(x)\sin(h)-\cos(x)(1-\cos(h))\over h}$

Rearranging the terms and the limit gives

$f'(x)=-\lim_{h\to 0}{\sin(x)\sin(h)\over h} - \lim_{h\to 0}{\cos(x)(1-\cos(h))\over h}$

Now because sin(x) and cos(x) do not vary as h varies, they can be taken out of the limit to give

$f'(x)=-\sin(x)\lim_{h\to 0}{\sin(h)\over h} - \cos(x)\lim_{h\to 0}{1-\cos(h)\over h}$

The value of the limits

$\lim_{h\to 0}{\sin(h)\over h} \quad\text{and}\quad \lim_{h\to 0}{1-\cos(h)\over h}$

are 1 and 0 respectively. Therefore, if f(x) = cos(x),

$f'(x)=-\sin(x). \,$

[edit] Proofs of derivatives of inverse trigonometric functions

The following derivatives are found by setting a variable y equal to the inverse trigonometric function that we wish to take the derivative of. Using implicit differentiation and then solving for dy/dx, the derivative of the inverse function is found in terms of y. To convert dy/dx back into being in terms of x, we can draw a reference triangle on the unit circle, letting theta be y. Using the Pythagorean theorem and the definition of the regular trigonometric functions, we can finally express dy/dx in terms of x.