Differentiation of Trigonometric Functions
The process of finding the derivatives of trigonometric functions is known as the differentiation of trigonometric functions. In other words, the differentiation of trigonometric functions is finding the rate of change of the function with respect to the variable. The six trigonometric functions have differentiation formulas that can be used in various application problems of the derivative.
The six basic trigonometric functions include the following: sine (sin x), cosine (cos x), tangent (tan x), cotangent (cot x), secant (sec x) and cosecant (cosec x). In this article, we will find the derivatives of the trigonometric functions and their proofs. Differentiation of trigonometric functions have applications in different fields such as electronics, computer programming and modeling different cyclic functions.
What is the Differentiation of Trigonometric Functions?
In trigonometry, differentiation of trigonometric functions is a mathematical process of determining the rate of change of the trigonometric functions with respect to the variable angle. The differentiation of trigonometric functions can be done using the derivatives of sin x and cos x by applying the quotient rule. The differentiation formulas of the six trigonometric functions are listed below:
 Derivation of sin x: (sin x)' = cos x
 Derivative of cos x: (cos x)' = sin x
 Derivative of tan x: (tan x)' = sec^{2} x
 Derivative of cot x: (cot x)' = cosec^{2} x
 Derivative of sec x: (sec x)' = sec x.tan x
 Derivative of cosec x: (cosec x)' = cosec x.cot x
Proof of Derivatives of Trigonometric Functions
Now, that we have the differentiation of trigonometric functions (sin x, cos x, tan x, cot x, sec x, cosec x), we will prove and derive the derivatives of trigonometric functions using various methods such as quotient rule, the first principle of differentiation, and chain rule along with some limit formulas.
Derivative of sin x
We will derive the derivative of sin x using the first principle of differentiation, that is, using the definition of limits. To derive the differentiation of the trigonometric function sin x, we will use the following limit and trigonometric formulas:
 sin (A+B) = sin A cos B + sin B cos A
 \(\lim_{x\rightarrow 0} \dfrac{\cos x 1}{x} = 0\)
 \(\lim_{x\rightarrow 0} \dfrac{\sin x}{x} = 1\)
Now, we will calculate the differentiation of trigonometric function sin x:
\(\begin{align}\frac{\mathrm{d} (\sin x)}{\mathrm{d} x} &= \lim_{h\rightarrow 0} \dfrac{\sin (x + h)\sin x}{(x+h)x} \\&= \lim_{h\rightarrow 0} \dfrac{\sin x \cos h +\cos x \sin h\sin x}{h}\\&=\lim_{h\rightarrow 0} \dfrac{\cos h 1 }{h}\sin x + \dfrac{\sin h}{h}\cos x\\&=(0)\sin x + (1)\cos x\\&=\cos x\end{align}\)
Therefore, d(sin x)/dx = cos x
Derivative of cos x
We will derive the derivative of cos x using the first principle of differentiation, that is, using the definition of limits. To derive the differentiation of the trigonometric function cos x, we will use the following limit and trigonometric formulas:
 cos (A + B) = cos A cos B  sin A sin B
 \(\lim_{x\rightarrow 0} \dfrac{\cos x 1}{x} = 0\)
 \(\lim_{x\rightarrow 0} \dfrac{\sin x}{x} = 1\)
Thus, we have
\(\begin{align}\frac{\mathrm{d} (\cos x)}{\mathrm{d} x} &= \lim_{h\rightarrow 0} \dfrac{\cos (x + h)\cos x}{(x+h)x} \\&= \lim_{h\rightarrow 0} \dfrac{\cos x \cos h \sin x \sin h\cos x}{h}\\&=\lim_{h\rightarrow 0} \dfrac{\cos h 1 }{h}\cos x  \dfrac{\sin h}{h}\sin x\\&=(0)\cos x  (1)\sin x\\&=\sin x\end{align}\)
Therefore, d(cos x)/dx = sin x
Derivative of tan x
We will determine the derivative of tan x using the quotient rule. We will use the following formulas and identities to calculate the derivative:
 (sin x)' = cos x
 (cos x)' = sin x
 tan x = sin x/ cos x
 cos^{2}x + sin^{2}x = 1
 sec x = 1/cos x
(tan x)' = (sin x/cos x)'
= [(sin x)' cos x  (cos x)' sin x]/cos^{2}x
= [cos x. cos x  (sin x). sin x]/cos^{2}x
= (cos^{2}x + sin^{2}x)/cos^{2}x
= 1/cos^{2}x
= sec^{2}x
Therefore, d(tan x)/dx = sec^{2}x
Derivative of cot x
We will determine the derivative of cot x using the quotient rule. We will use the following formulas and identities to calculate the derivative:
 (sin x)' = cos x
 (cos x)' = sin x
 cot x = cos x/ sin x
 cos^{2}x + sin^{2}x = 1
 cosec x = 1/sin x
(cot x)' = (cos x/sin x)'
= [(cos x)' sin x  (sin x)' cos x]/sin^{2}x
= [sin x. sin x  cos x. cos x]/sin^{2}x
= (sin^{2}x  cos^{2}x)/sin^{2}x
= 1/sin^{2}x
= cosec^{2}x
Therefore, d(cot x)/dx = cosec^{2}x
Derivative of sec x
We will determine the derivative of sec x using the chain rule. We will use the following formulas and identities to calculate the derivative:
 sec x = 1/cos x
 tan x = sin x/cos x
 (cos x)' = sin x
(sec x)' = (1/cos x)' = (1/cos^{2}x).(cos x)'
= (1/cos^{2}x).(sin x)
= sin x/cos^{2}x
= (sin x/cos x).(1/cos x)
= tan x sec x
Therefore, d(sec x)/dx = tan x sec x
Derivative of cosec x
We will determine the derivative of cosec x using the chain rule. We will use the following formulas and identities to calculate the derivative:
 cosec x = 1/sin x
 cot x = cos x/sin x
 (sin x)' = cos x
(cosec x)' = (1/sin x)' = (1/sin^{2}x).(sin x)'
= (1/sin^{2}x).(cos x)
= cos x/sin^{2}x
= (cos x/sin x).(1/sin x)
= cot x cosec x
Therefore, d(cosec x)/dx = cot x cosec x
Applications of Differentiation of Trigonometric Functions
The differentiation of trigonometric functions has various applications in the field of mathematics and real life. Some of them are listed below:
 It is used to determine the slope of the tangent line to a trigonometric curve y = f(x).
 It is used to determine the slope of the normal line to a trigonometric curve y = f(x).
 It helps to determine the equation of the tangent line or normal line of a curve.
 Differentiation of trigonometric functions have applications in different fields such as electronics, computer programming and modeling different cyclic functions.
 We use the derivatives of trigonometric functions to determine the maximum and minimum values of particular functions.
Differentiation of Inverse Trigonometric Functions
The differentiation of inverse trigonometric functions is done by setting the function equal to y and applying implicit differentiation. Let us list the derivatives of the inverse trigonometric functions along with their domains (arcsin x, arccos x, arctan x, arccot x, arcsec x, arccosec x):
 (arcsin x)' = 1/√(1  x^{2}) , 1 < x < 1
 (arccos x)' = 1/√(1  x^{2}) , 1 < x < 1
 (arctan x)' = 1/(1 + x^{2}) , ∞ < x < ∞
 (arccot x)' = 1/(1 + x^{2}) , ∞ < x < ∞
 (arcsec x)' = 1/x√(x^{2 } 1) , x ∈ (∞, 1) ∪ (1, ∞)
 (arccosec x)' = 1/x√(x^{2 } 1) , x ∈ (∞, 1) ∪ (1, ∞)
AntiDifferentiation of Trigonometric Functions
Anti differentiation of trigonometric functions is the reverse process of differentiation of trigonometric functions. This process is also called the integration of trigonometric functions. The list of antiderivatives of the trigonometric functions is given below as:
 ∫ sinx dx = cos x + C
 ∫ cosx dx = sin x + C
 ∫ tanx dx = ln sec x + C
 ∫ cotx dx = ln sin x + C
 ∫ secx dx = ln sec x + tan x + C
 ∫ cosecx dx = ln cosec x + cot x + C
Here C is the constant of integration.
Related Topics on Derivatives of Trigonometric Functions
Important Notes on Differentiation of Trigonometric Functions
 Derivation of sin x: (sin x)' = cos x
 Derivative of cos x: (cos x)' = sin x
 Derivative of tan x: (tan x)' = sec^{2} x
 Derivative of cot x: (cot x)' = cosec^{2} x
 Derivative of sec x: (sec x)' = sec x.tan x
 Derivative of cosec x: (cosec x)' = cosec x.cot x
Differentiation of Trigonometric Functions Examples

Example 1: Show the differentiation of trigonometric function cos x using chain rule.
Solution: The chain rule for differentiation is: (f(g(x)))’ = f’(g(x)) . g’(x). Now, to evaluate the derivative of cos x using the chain rule, we will use certain trigonometric properties and identities such as:
 \(\cos (\dfrac{\pi}{2}  \theta) = \sin \theta\)
 \(\sin (\dfrac{\pi}{2}  \theta) = \cos \theta\)
 d(sin x)/dx = cos x
Using the above three trigonometric properties, we can write the derivative of cos x as the derivative of sin (π/2  x), that is, d(cos x)/dx = d (sin (π/2  x))/dx . Using chain rule, we have,
\(\begin{align} \frac{\mathrm{d} \cos x}{\mathrm{d} x} &=\frac{\mathrm{d} \sin(\dfrac{\pi}{2}x)}{\mathrm{d} x}\\&=\cos(\dfrac{\pi}{2}x).(1)\\&=\cos(\dfrac{\pi}{2}x)\\&= \sin x\end{align}\)
Answer: Hence, we have derived the derivative of cos x as sin x using chain rule.

Example 2: Show the differentiation of trigonometric function tan x by the first principle of differentiation.
Solution: We will derive the derivative of tan x using the first principle of differentiation, that is, using the definition of limits. To derive the differentiation of the trigonometric function tan x, we will use the following limit and trigonometric formulas:
 tan (A + B) = (tan A + tan B)/(1  tan A tan B)
 \(\lim_{x\rightarrow 0} \tan x = 0\)
 \(\lim_{x\rightarrow 0} \dfrac{\tan x}{x} = 1\)
 tan x = sin x/ cos x
 1 + tan^{2}x = sec^{2}x
\(\begin{align}\frac{\mathrm{d} (\tan x)}{\mathrm{d} x} &= \lim_{h\rightarrow 0} \dfrac{\tan (x + h)\tan x}{(x+h)x} \\&= \lim_{h\rightarrow 0} \dfrac{\frac{\tan x +\tan h}{1\tan x \tan h}\tan x}{h}\\&=\lim_{h\rightarrow 0} \dfrac{\tan x +\tan h \tan x+\tan^2 x \tan h }{h(1\tan x \tan h)}\\&=\lim_{h\rightarrow 0} \dfrac{\tan h}{h}\times \lim_{h\rightarrow 0} \dfrac{1+\tan^2 x}{(1\tan x \tan h)} \\&=1 \times \dfrac{\sec^2x}{1}\\&=\sec^2x\end{align}\)
Answer: Hence, we have derived the derivative of tan x as sec^{2}x using the first principle of differentiation.
FAQs on Differentiation of Trigonometric Functions
What is the Differentiation of Trigonometric Functions in Trigonometry?
In trigonometry, differentiation of trigonometric functions is a mathematical process of determining the rate of change of the trigonometric functions with respect to the angle variable. The process of finding the derivatives of circular trigonometric functions is known as the differentiation of trigonometric functions.
What are the Derivatives of the 6 Trig Functions?
The differentiation formulas of the six trigonometric functions are listed below:
 Derivation of sin x: (sin x)' = cos x
 Derivative of cos x: (cos x)' = sin x
 Derivative of tan x: (tan x)' = sec^{2} x
 Derivative of cot x: (cot x)' = cosec^{2} x
 Derivative of sec x: (sec x)' = sec x.tan x
 Derivative of cosec x: (cosec x)' = cosec x.cot x
What are the Applications of Differentiation of Trigonometric Functions?
The differentiation of trigonometric functions has various applications in the field of mathematics and real life.
 It helps to determine the equation of the tangent line or normal line of a curve.
 Differentiation of trigonometric functions have applications in different fields such as electronics, computer programming and modeling different cyclic functions.
 We use the differentiation of a trigonometric function to determine the maximum and minimum values of particular functions.
What is the AntiDifferentiation of Trigonometric Functions in Trigonometry?
Anti differentiation of trigonometric functions is the reverse process of differentiation of trigonometric functions. This process is also called the integration of trigonometric functions.
What are the AntiDerivatives of the Six Trigonometric Functions?
The list of antiderivatives of the trigonometric functions is given below as:
 ∫ sinx dx = cos x + C
 ∫ cosx dx = sin x + C
 ∫ tanx dx = ln sec x + C
 ∫ cotx dx = ln sin x + C
 ∫ secx dx = ln sec x + tan x + C
 ∫ cosecx dx = ln cosec x + cot x + C
What are the Differentiation Formulas for Finding the Derivative of Trigonometric Functions?
Derivative of Trigonometric Functions can be calculated using various methods such as quotient rule, the first principle of differentiation, and chain rule along with some limit formulas.
 (sin x)' = cos x
 (cos x)' = sin x
 (tan x)' = sec^{2} x
 (cot x)' = cosec^{2} x
 (sec x)' = sec x.tan x
 (cosec x)' = cosec x.cot x
How Do You Derive the Derivatives of Trigonometric Functions?
The differentiation of trigonometric functions can be done using different methods of differentiation such as first principle of derivatives, product rule, quotient rule and chain rule.
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