Differentiation of cosecx

Here you will learn what is the differentiation of cosecx and its proof by using first principle.

Let’s begin –

Differentiation of cosecx

The differentiation of cosecx with respect to x is -cosecx.cotx

i.e. $d\over dx$ (cosecx) = -cosecx.cotx

Proof Using First Principle :

Let f(x) = cosec x. Then, f(x + h) = cosec(x + h)

$\therefore$   $d\over dx$(f(x)) = $lim_{h\to 0}$ $f(x + h) – f(x)\over h$

$d\over dx$(f(x)) = $lim_{h\to 0}$ $cosec(x + h) – cosec x\over h$

$\implies$  $d\over dx$(f(x)) = $lim_{h\to 0}$ ${1\over sin(x + h)} – {1\over sin x}\over h$

$\implies$  $d\over dx$(f(x)) = $lim_{h\to 0}$ $sin x – sin(x + h)\over h sin x sin(x +h)$

By using trigonometry formula,

[sin C – sin D = $2sin ({C – D\over 2})cos ({C + D\over 2})$]

$\implies$ $d\over dx$(f(x)) = $lim_{h\to 0}$ $2sin ({x – x – h\over 2})cos({x + x + h\over 2})\over h sin x sin (x + h)$

$\implies$ $d\over dx$(f(x)) = $lim_{h\to 0}$ $2sin ({-h\over 2})cos({x + h/2})\over h sin x sin (x + h)$

$\implies$ $d\over dx$(f(x)) = -$lim_{h\to 0}$ $cos ({x + h/2})\over sin x sin(x + h)$.$lim_{h\to 0}$ $sin(h/2)\over (h/2)$

because, [$lim_{h\to 0}$$sin(h/2)\over (h/2)$ = 1]

$\implies$ $d\over dx$(f(x)) = -$cos x\over sin x sin x$(1) = -cot x cosec x

Hence, $d\over dx$ (cosec x) = =cosecx.cotx

Example : What is the differentiation of cosec x + x with respect to x?

Solution : Let y = cosec x + x

$d\over dx$(y) = $d\over dx$(cosec x + x)

$\implies$ $d\over dx$(y) = $d\over dx$(cosec x) + $d\over dx$(x)

By using cosecx differentiation we get,

$\implies$ $d\over dx$(y) = -cosec x cot x + 1

Hence, $d\over dx$(sec x + x) = -cosec x cot x + 1

Example : What is the differentiation of $cosec\sqrt{x}$ with respect to x?

Solution : Let y = $cosec\sqrt{x}$

$d\over dx$(y) = $d\over dx$($cosec\sqrt{x}$)

By using chain rule we get,

$\implies$ $d\over dx$(y) = $1\over 2\sqrt{x}$($-cosec \sqrt{x}.cot\sqrt{x}$)

Hence, $d\over dx$($cosec\sqrt{x}$) = -$1\over 2\sqrt{x}$($cosec \sqrt{x}.cot\sqrt{x}$)

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