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The concept of derivatives of trigonometric functions is an essential part of calculus. In this article, we will provide an overview of the derivatives of trigonometric functions, along with examples, solutions, and frequently asked questions.
Derivatives of Trigonometric Functions
The derivatives of the six trigonometric functions (sine, cosine, tangent, cotangent, secant, and cosecant) can be calculated using the following formulas:
$\frac {d}{dx} (sin x) = cos x $
$\frac {d}{dx} (cos x) = -sin x $
$\frac {d}{dx} (tan x) = sec^2x $
$\frac {d}{dx} (cot x) = -cosec^2x $
$\frac {d}{dx} (sec x) = sec x . tan x $
$\frac {d}{dx} (cosec x) = -cosec x . cot x$
Important formula to find the derivative
$\frac {d}{dx} f(x) =\displaystyle \lim_{h \to 0} \frac {f (x+h) – f(x)}{h}$
$\displaystyle \lim_{x \to 0} \frac {sinx}{x}=1$
Proof of sin derivative
$\frac {d}{dx} (sin x) = \displaystyle \lim_{h \to 0} \frac {sin (x+h) – sin x}{h}= \displaystyle \lim_{h \to 0} \frac {2cos (\frac {2x+h}{2}) sin \frac {h}{2}}{h}$
$=\displaystyle \lim_{h \to 0} cos (x + \frac {h}{2}) \displaystyle \lim_{h \to 0} \frac {sin \frac {h}{2}}{\frac {h}{2}} = cos x$
as $\displaystyle \lim_{x \to 0} \frac {sinx}{x}=1$
Proof of cos derivative
$\frac {d}{dx} (cos x) = \displaystyle \lim_{h \to 0} \frac {cos (x+h) – cos x}{h}= \displaystyle \lim_{h \to 0} -\frac {2sin (\frac {2x+h}{2}) sin \frac {h}{2}}{h}$
$=-\displaystyle \lim_{h \to 0} sin (x + \frac {h}{2}) \displaystyle \lim_{h \to 0} \frac {sin \frac {h}{2}}{\frac {h}{2}} = -sin x$
as $\displaystyle \lim_{x \to 0} \frac {sinx}{x}=1$
Proof of tan derivative
$\frac {d}{dx} (tan x) = \displaystyle \lim_{h \to 0} \frac {tan (x+h) – tan x}{h}= \displaystyle \lim_{h \to 0} \frac {1}{h}[\frac {sin(x+h)}{cos(x+h) }- \frac {sin(x)}{cos(x)}]$
$=\displaystyle \lim_{h \to 0} \frac {1}{h}\frac {sin(x+h)cos (x) – sin(x) cos (x+h)}{cos (x+h) cos x}$
$=\displaystyle \lim_{h \to 0} \frac {1}{h} \frac {sin (x+h -x)}{cos (x+h) cos x} $
$=\displaystyle \lim_{h \to 0} \frac {sin h}{h} \displaystyle \lim_{h \to 0} \frac {1}{cos (x+h) cos x}= sec^2 x$
as $\displaystyle \lim_{x \to 0} \frac {sinx}{x}=1$
Examples
Example 1
Find the derivative of f(x) = sin 2x.
Solution
Using the chain rule of differentiation, we get
$\frac {d}{dx}(sin 2x) = cos 2x \times 2 = 2cos 2x
or
we can find using product rule of differentiation, which states that the derivative of the product of two functions is the first function times the derivative of the second function, plus the second function times the derivative of the first function.
$\frac {d}{dx}(sin 2x) = \frac {d}{dx} (2sinx cos x) =2 \frac {d}{dx} (sin x cos x)$
$ =2 [ cos x \frac {d}{dx} (sinx ) + sin (x) \frac {d}{dx} (cos x)]= 2 (cos^2 x – sin^2 x) = 2 cos 2x$
Example 2
Find the derivative of g(x) = cot x.
Solution
Using the formula for the derivative of cotangent, we get:
$\frac {d}{dx}(cot x) = -cosec^2x $
Example 3
Find the derivative of h(x) = cos 3x.
Solution
$\frac {d}{dx}(cos 3x) = -3sin 3x.$
Example 4
Find the derivative of f(x) = 2tan x + sin 2x.
Solution
$\frac {d}{dx}(2tan x + sin 2x) = 2sec^2x + 2cos 2x$
Derivatives of trigonometric functions are an important part of calculus and are included in the CBSE board syllabus. By using the formulas and rules of differentiation, students can calculate the derivatives of the six trigonometric functions. Practicing examples will help students master this topic.
