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 ?
