Question

Find the derivative ofcosec by using first principle of derivative

Answer 1

Given: $\rm f(x) = cosec(x)$

To find: derivative of given function

Solution:

Using the definition of first principles we have,

$\tt f \prime(x) = \lim \limits_{h \to0} \dfrac{f(x + h) - f(x)}{h}$

$\tt \implies f \prime(x) = \lim \limits_{h \to0} \dfrac{ cosec(x + h) - cosec(x)}{h}$

$\tt \implies f \prime(x) = \lim \limits_{h \to0} \dfrac{ \frac{1}{sin(x + h)} - \frac1{sin(x)}}{h}$

$\tt \implies f \prime(x) = \lim \limits_{h \to0} \dfrac{sin(x) - sin(x + h)}{h \: sin(x + h)sin(x)}$

Now using the identity:-

• sin(A) - sin(B) = 2sin(A-B/2) cos(A+B/2)

$\tt \implies f \prime(x) = \lim \limits_{h \to0} \dfrac{2 \: sin( \frac{x - x - h}{2}) \cdot \: cos( \frac{x + x + h}{2})}{h \: sin(x + h)sin(x)}$

$\tt \implies f \prime(x) = \lim \limits_{h \to0} \dfrac{2 \: sin( \frac{ - h}{2}) \cdot \: cos( x + \frac{h}{2})}{h \: sin(x + h)sin(x)}$

$\tt \implies f \prime(x) = \lim \limits_{h \to0} \dfrac{ - 2 \: sin( \frac{h}{2}) \cdot \: cos( x + \frac{h}{2})}{h \: sin(x + h)sin(x)}$

$\tt \implies f \prime(x) = \lim \limits_{h \to0} \dfrac{ - \not2 \: sin( \frac{h}{2}) \cdot \: cos( x + \frac{h}{2})}{ \not 2 \cdot \frac{h}{2} \: sin(x + h)sin(x)}$

$\tt \implies f \prime(x) =\lim \limits_{h \to0} \dfrac{sin( \frac{h}{2} )}{ (\frac{h}{2}) } \cdot \lim \limits_{h \to0} \dfrac{ - cos( x + \frac{h}{2})}{ sin(x + h)sin(x)}$

Using the standard limit:-

• lim(x → 0) sinx/x = 1

$\tt \implies f \prime(x) =\lim \limits_{h \to0} \dfrac{ - cos( x + \frac{h}{2})}{ sin(x + h)sin(x)}$

$\tt \implies f \prime(x) = \dfrac{ - cos( x )}{ sin(x)sin(x)}$

$\tt \implies f \prime(x) =- cosec( x ) \: cot(x)$

Hence the derivative of cosec(x) is - cosec(x) cot(x).

Answer 2

Hence, we have derived the derivative of cosec x to be -cot x cosec x using the first principle of differentiation.

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Derivative of Cosec x Proof By First Principle

cosec x = 1/sin x.

limh→0(sin(x+h)−sinx)h=cosx lim h → 0 ( sin ⁡ ( x + h ) − sin ⁡ x ) h = cos ⁡ x.

cot x = cos x/sin x.