Differentiation of cotx

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

Let’s begin –

Differentiation of cotx

The differentiation of cotx with respect to x is \(-cosec^2x\).

i.e. \(d\over dx\) (cotx) = \(-cosec^2x\)

Proof Using First Principle :

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

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

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

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

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

By using trigonometry formula,

[sin A cos B – cos A sin B = sin (A – B)]

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

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

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

\(\implies\) \(d\over dx\)(f(x)) = -1.\(1\over sin x sin x\) = \(-cosec^2x\)

Hence, \(d\over dx\) (cot x) = \(-cosec^2x\)

Example : What is the differentiation of cot x + 1 with respect to x?

Solution : Let y = cot x + 1

\(d\over dx\)(y) = \(d\over dx\)(cot x + 1)

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

By using cotx differentiation we get,

\(\implies\) \(d\over dx\)(y) = \(-cosec^2x\) + 0

Hence, \(d\over dx\)(cot x + 1) = \(-cosec^2x\) 

Example : What is the differentiation of \(cot\sqrt{x}\) with respect to x?

Solution : Let y = \(cot\sqrt{x}\)

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

By using chain rule we get,

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

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


Related Questions

What is the Differentiation of cot inverse x ?

What is the Differentiation of cosx ?

What is the Integration of Cot x ?

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