Mathematical process of finding the derivative of a trigonometric function
The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. For example, the derivative of the sine function is written sin′(a) = cos(a), meaning that the rate of change of sin(x) at a particular angle x = a is given by the cosine of that angle.
All derivatives of circular trigonometric functions can be found from those of sin(x) and cos(x) by means of the quotient rule applied to functions such as tan(x) = sin(x)/cos(x). Knowing these derivatives, the derivatives of the inverse trigonometric functions are found using implicit differentiation.
Proofs of derivatives of trigonometric functions
Limit of sin(θ)/θ as θ tends to 0
The diagram at right shows a circle with centre O and radius r = 1. Let two radii OA and OB make an arc of θ radians. Since we are considering the limit as θ tends to zero, we may assume θ is a small positive number, say 0 < θ < 1/2 π in the first quadrant.
In the diagram, let R1 be the triangle OAB, R2 the circular sectorOAB, and R3 the triangle OAC.
Since each region is contained in the next, one has:
Moreover, since sin θ > 0 in the first quadrant, we may divide through by 1/2sin θ, giving:
In the last step we took the reciprocals of the three positive terms, reversing the inequities.
We conclude that for 0 < θ < 1/2 π, the quantity sin(θ)/θ is always less than 1 and always greater than cos(θ). Thus, as θ gets closer to 0, sin(θ)/θ is "squeezed" between a ceiling at height 1 and a floor at height cos θ, which rises towards 1; hence sin(θ)/θ must tend to 1 as θ tends to 0 from the positive side:
For the case where θ is a small negative number –1/2 π < θ < 0, we use the fact that sine is an odd function:
Limit of (cos(θ)-1)/θ as θ tends to 0
The last section enables us to calculate this new limit relatively easily. This is done by employing a simple trick. In this calculation, the sign of θ is unimportant.
Using cos2θ – 1 = –sin2θ,the fact that the limit of a product is the product of limits, and the limit result from the previous section, we find that:
Limit of tan(θ)/θ as θ tends to 0
Using the limit for the sine function, the fact that the tangent function is odd, and the fact that the limit of a product is the product of limits, we find:
Proofs of derivatives of inverse trigonometric functions
The following derivatives are found by setting a variabley 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 θ be y. Using the Pythagorean theorem and the definition of the regular trigonometric functions, we can finally express dy/dx in terms of x.
Differentiating the inverse sine function
We let
Where
Then
Taking the derivative with respect to on both sides and solving for dy/dx:
Substituting in from above,
Substituting in from above,
Differentiating the inverse cosine function
We let
Where
Then
Taking the derivative with respect to on both sides and solving for dy/dx:
Substituting in from above, we get
Substituting in from above, we get
Alternatively, once the derivative of is established, the derivative of follows immediately by differentiating the identity so that .
Differentiating the inverse tangent function
We let
Where
Then
Taking the derivative with respect to on both sides and solving for dy/dx:
Left side:
using the Pythagorean identity
Right side:
Therefore,
Substituting in from above, we get
Differentiating the inverse cotangent function
We let
where . Then
Taking the derivative with respect to on both sides and solving for dy/dx:
Left side:
using the Pythagorean identity
Right side:
Therefore,
Substituting ,
Alternatively, as the derivative of is derived as shown above, then using the identity follows immediately that
Differentiating the inverse secant function
Using implicit differentiation
Let
Then
(The absolute value in the expression is necessary as the product of secant and tangent in the interval of y is always nonnegative, while the radical is always nonnegative by definition of the principal square root, so the remaining factor must also be nonnegative, which is achieved by using the absolute value of x.)
Using the chain rule
Alternatively, the derivative of arcsecant may be derived from the derivative of arccosine using the chain rule.
Let
Where
and
Then, applying the chain rule to :
Differentiating the inverse cosecant function
Using implicit differentiation
Let
Then
(The absolute value in the expression is necessary as the product of cosecant and cotangent in the interval of y is always nonnegative, while the radical is always nonnegative by definition of the principal square root, so the remaining factor must also be nonnegative, which is achieved by using the absolute value of x.)
Using the chain rule
Alternatively, the derivative of arccosecant may be derived from the derivative of arcsine using the chain rule.