Derivatives of Trigonometric Functions

Last Updated : 1 Nov, 2025

Derivatives of the trigonometric functions are first calculated through the limit definition. Before calculating these derivatives and looking at their proofs, it is necessary to revisit some identities of limit related to trigonometric functions. 

\lim_{x \to 0}\frac{sin(x)}{x}

 and 

\lim_{x \to 0}\frac{cos(x) - 1}{x}

Let's calculate the derivatives for the three most important trigonometric functions. 

  1. \frac{d(sin(x))}{dx} = cos(x)
  2. \frac{d(cos(x))}{dx} = -sin(x)
  3. \frac{d(tan(x))}{dx} = sec^2(x)

Proving the derivative for the sine function \frac{d(sin(x))}{dx} = cos(x)

Using the previously mentioned definition of the derivatives 

\frac{df(x)}{dx} = \lim_{h \to 0}\frac{f(x + h) - f(x) }{h} \\ = \frac{dsin(x)}{dx} = \lim_{h \to 0}\frac{sin(x + h) - sin(x) }{h}

⇒\frac{dsin(x)}{dx} = \lim_{h \to 0}\frac{sin(x)cos(h) + sin(h)cos(x) - sin(x) }{h}

⇒\frac{dsin(x)}{dx} = \lim_{h \to 0}\frac{sin(x)(cos(h)-1) + sin(h)cos(x)}{h} \\ = \frac{dsin(x)}{dx} = \lim_{h \to 0}\frac{sin(x)(cos(h)-1)}{h}+ \lim_{h \to 0}\frac{sin(h)cos(x)}{h} \\ = \frac{dsin(x)}{dx} = sin(x)\lim_{h \to 0}\frac{(cos(h)-1)}{h}+ cos(x)\lim_{h \to 0}\frac{sin(h)}{h} \\

Using the limit identities described above, 

⇒\frac{d(sin(x))}{dx} = cos(x)

Proving the derivative for the cosine function \frac{d(cos(x))}{dx} = -sin(x)

Using the previously mentioned definition of the derivatives 

\frac{df(x)}{dx} = \lim_{h \to 0}\frac{f(x + h) - f(x) }{h}

⇒\frac{df(x)}{dx} = \lim_{h \to 0}\frac{cos(x + h) - cos(x) }{h}

⇒\frac{df(x)}{dx} = \lim_{h \to 0}\frac{cos(x)cos(h) - sin(x)sin(h) -  cos(x) }{h}

⇒\frac{df(x)}{dx} = \lim_{h \to 0}\frac{cos(x)(cos(h) -1)  -sin(x)sin(h)}{h}

⇒\frac{df(x)}{dx} = cos(x)\lim_{h \to 0}\frac{(cos(h) -1)}{h} -sin(x)\lim_{h \to 0}\frac{sin(h)}{h}

Using the limit identities described above, 

⇒\frac{d(cos(x))}{dx} = -sin(x)

Proving the derivative for the tangent function \frac{d(tan(x))}{dx} = sec^2(x)

Now that we have the derivatives for the sine and cosine functions. Derivatives of other functions can be calculated simply through Quotient and product rules. 

Quotient rule says, for a function f(x) = \frac{h(x)}{g(x)}, the derivative of this function is given by, 

f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{(g(x))^2}

In this case tan(x) = \frac{sin(x)}{cos(x)}

Thus, h(x) = sin(x) and g(x) = cos(x). 

f'(x) = \frac{cos(x)\frac{dsin(x)}{dx} - sin(x)\frac{dcos(x)}{dx} }{(cos(x))^2}

⇒f'(x) = \frac{cos(x)cos(x) + sin(x)sin(x) }{(cos(x))^2}

⇒f'(x) = \frac{1 }{(cos(x))^2}

⇒\frac{d(tan(x))}{dx} = sec^2(x)

Proving the derivative for the secant function \frac{d(sec(x))}{dx} = sec(x)tan(x)

This can be proved easily through chain rule, 

\frac{d(sec(x))}{dx} = \frac{d(\frac{1}{cos(x)})}{dx}

⇒\frac{d(sec(x))}{dx} = \frac{-1}{cos^2(x)}\frac{d(cos(x))}{dx}

⇒\frac{d(sec(x))}{dx} = \frac{1}{cos^2(x)}sin(x)

⇒\frac{d(sec(x))}{dx} = sec(x)tan(x)

Proving the derivative for the co-secant function \frac{d(cosec(x))}{dx} = -cosec(x)cot(x)

This can be proved easily through chain rule, 

\frac{d(cosec(x))}{dx} = \frac{d(\frac{1}{sin(x)})}{dx}

⇒\frac{d(cosec(x))}{dx} = \frac{-1}{sin^2(x)}\frac{d(sin(x))}{dx}

⇒\frac{d(cosec(x))}{dx} = \frac{-1}{sin^2(x)}cos(x)

⇒\frac{d(cosec(x))}{dx} = -cosec(x)cot(x)

Proof of Product Rule

There are different ways to prove the product rule which includes definition of the derivative and logarithmic differentiation, lets look at the latter way, y = f(x)g(x)

This can easily be proved by taking the natural log,

ln y = ln{f(x)g(x)} = ln f(x) + ln g(x)

y'/y= f'(x)/f(x)+ g'(x)/g(x)

y'=y[\frac{f'(x)g(x)+ g'(x)f(x)}{f(x)g(x)}]\\ y'=f(x)g(x)[\frac{f'(x)g(x)+ g'(x)f(x)}{f(x)g(x)}]\\ y'= f'(x)g(x)+ g'(x)f(x)

Hence, proved.

Power rule of Derivatives

This is the most commonly used rule in derivatives. It says, \frac{dx^n}{dx} = nx^{n - 1}

This can be derived through the limit definition of the derivatives. 

\frac{d(x^n)}{dx} = \lim_{h \to 0}\frac{(x + h)^n - x^n}{h}

(x + h)n can be opened through binomial expansion, 

(x + h)^n = x^n + nx^{n-1}h + \frac{n(n-1)}{2!}x^{n-2}h^2 + ....

⇒\frac{d(x^n)}{dx} = \lim_{h \to 0}\frac{(x^n + \frac{n!}{(n-1)!}x^{n-1}h + \frac{n!}{(n-2)!2!}h^2 + .... - x^n)}{h}

⇒\frac{d(x^n)}{dx} = \lim_{h \to 0}\frac{\frac{n!}{(n-1)!}x^{n-1}h + \frac{n!}{(n-2)!2!}h^2 + ....)}{h}

⇒\frac{d(x^n)}{dx} =\frac{n!}{(n-1)!}x^{n-1} +  \lim_{h \to 0}\frac{( \frac{n!}{(n-2)!2!}h^2 + ....)}{h}

⇒\frac{d(x^n)}{dx} =nx^{n-1}

Example: Find the derivative of f(x) = √x. 

Solution: 

\frac{d(\sqrt{x})}{dx} = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}

⇒\frac{d(\sqrt{x})}{dx} = \lim_{h \to 0}\frac{\sqrt{x + h} - \sqrt{x}}{h}

This is indeterminate form, using rationalization 

⇒\frac{d(\sqrt{x})}{dx} = \lim_{h \to 0}\frac{(\sqrt{x + h} - \sqrt{x})(\sqrt{x + h} + \sqrt{x})}{h(\sqrt{x + h} + \sqrt{x})}

⇒\frac{d(\sqrt{x})}{dx} = \lim_{h \to 0} \frac{(x + h) -x}{h(\sqrt{x + h} + \sqrt{x})}

⇒\frac{d(\sqrt{x})}{dx} = \lim_{h \to 0} \frac{h}{h(\sqrt{x + h} + \sqrt{x})}

⇒\frac{d(\sqrt{x})}{dx} = \lim_{h \to 0} \frac{1}{(\sqrt{x + h} + \sqrt{x})}

⇒\frac{d(\sqrt{x})}{dx} = \frac{1}{2\sqrt{x}}

Applications of Differentiation of Trigonometric Functions

  • This equation gives the rate of change in a system that is oscillating or vibrating. Examples would include sound waves, pendulums, and alternating currents.
  • It helps in finding the equation of the tangent or normal line to a curve.
  • Differentiation of the trigonometric functions has many uses in electrical engineering, computer programming, and modeling cyclic or wave-like phenomena.
  • It helps in finding the maximum and minimum values of the trigonometric functions.
  • It is used in finding the slope of a tangent line to a trigonometric curve y = f(x).
  • It is used to find the slope of the normal line to a trigonometric curve y = f(x).

Derivatives of Trigonometric Functions - Sample Problems

Question 1: Find the derivative for the function f(x) at x = 0. 

f(x) = sin2(x)

Solution: 

This derivative is the combination of power rule and chain rule. 

f'(x) = \frac{dsin^2x}{dx}

⇒f'(x) = 2sin(x)\frac{dsinx}{dx}

Using the previous results for the derivative of sin(x). 

⇒f'(x) = 2sin(x)cos(x)

At x = 0 

f'(x) = 0

Question 2: Find the derivative for the function f(x) at x = 0. 

f(x) = 5sec(x) + 2cos(x)

Solution: 

This derivative is simple

f'(x) = \frac{d(5sec(x) + 2cos(x))}{dx}

⇒f'(x) = \frac{d(5sec(x))}{dx} + \frac{d(2cos(x))}{dx}

Using the previous results for the derivative of sin(x). 

⇒f'(x) = 5sec(x)tan(x) - 2sin(x). 

at x = 0 

f'(x) = 0

Question 3: Find the derivative for the function f(x) at x = 1. f(x) = x2 + x4

Solution: 

This derivative is simple application of power rule 

f'(x) = \frac{d(x^2  + x^4)}{dx}

⇒f'(x) = \frac{d(x^2)}{dx} + \frac{d(x^4)}{dx}

Using the previous results for the derivative of sin(x). 

⇒f'(x) = 2x + 4x3

At x = 1 

f'(x) = 6

Question 4: Find the derivative for the function f(x). 

f(x) = \frac{e^x + 1}{x}

Solution: 

This derivative is simple application of quotient rule.

Quotient rule says, for a function f(x) = \frac{h(x)}{g(x)}, the derivative of this function is given by, 

f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{(g(x))^2}

Here h(x) = ex + 1 and g(x) = x

f'(x) = \frac{x\frac{d(e^x + 1)}{dx} - (e^x + 1)\frac{dx}{dx}}{x^2}

⇒f'(x) = \frac{xe^x - (e^x + 1)}{x^2}

Question 5: Use the Product rule to differentiate the given function,

y = 5xsinx + 4x2cosx

Solution:

Differentiating using Product Rule,

y'= 5sinx + 5xcosx + 8xcosx - 4x2sinx

y'= 5sinx + 13xcosx - 4x2sinx

Practice Problems on Derivatives of Trigonometric Functions

1. Find the derivative of f(x) = 3sin(2x) + 4cos(x).

2. Differentiate g(x) = tan(x²) using the chain rule.

3. Find the derivative of h(x) = x · sin(x) using the product rule.

4. Determine the derivative of f(x) = sec(3x) - csc(x).

5. Calculate the second derivative of g(x) = cos(2x).

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