Here you will learn what is the differentiation of cotx and its proof by using first principle.<\/p>\n
Let’s begin –<\/p>\n
\nThe differentiation of cotx with respect to x is \\(-cosec^2x\\).<\/p>\n
i.e. \\(d\\over dx\\) (cotx) = \\(-cosec^2x\\)<\/p>\n<\/blockquote>\n
Proof Using First Principle :<\/h2>\n
\nLet f(x) = cot x. Then, f(x + h) = cot(x + h)<\/p>\n
\\(\\therefore\\)\u00a0 \u00a0\\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\(f(x + h) – f(x)\\over h\\)<\/p>\n
\\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\(cot(x + h) – cot x\\over h\\)<\/p>\n
\\(\\implies\\)\u00a0 \\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\({cos(x + h)\\over sin(x + h)} – {cos x\\over sin x}\\over h\\)<\/p>\n
\\(\\implies\\)\u00a0 \\(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)\\)<\/p>\n
By using trigonometry formula,<\/p>\n
[sin A cos B – cos A sin B = sin (A – B)]<\/p>\n
\\(\\implies\\) \\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\(sin h\\over h\\).\\(1\\over sin x sin (x + h)\\)<\/p>\n
\\(\\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)\\)<\/p>\n
because, [\\(lim_{h\\to 0}\\)\\(sin(h\/2)\\over (h\/2)\\) = 1]<\/p>\n
\\(\\implies\\) \\(d\\over dx\\)(f(x)) = -1.\\(1\\over sin x sin x\\) = \\(-cosec^2x\\)<\/p>\n
Hence, \\(d\\over dx\\) (cot x) = \\(-cosec^2x\\)<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : What is the differentiation of cot x + 1 with respect to x?<\/p>\n
Solution<\/strong><\/span> : Let y = cot x + 1<\/p>\n
\\(d\\over dx\\)(y) = \\(d\\over dx\\)(cot x + 1)<\/p>\n
\\(\\implies\\) \\(d\\over dx\\)(y) = \\(d\\over dx\\)(cot x) + \\(d\\over dx\\)(1)<\/p>\n
By using cotx differentiation we get,<\/p>\n
\\(\\implies\\) \\(d\\over dx\\)(y) = \\(-cosec^2x\\) + 0<\/p>\n
Hence, \\(d\\over dx\\)(cot x + 1) = \\(-cosec^2x\\)\u00a0<\/p>\n
Example<\/strong><\/span> : What is the differentiation of \\(cot\\sqrt{x}\\) with respect to x?<\/p>\n
Solution<\/strong><\/span> : Let y = \\(cot\\sqrt{x}\\)<\/p>\n
\\(d\\over dx\\)(y) = \\(d\\over dx\\)(\\(cot\\sqrt{x}\\))<\/p>\n
By using chain rule we get,<\/p>\n
\\(\\implies\\) \\(d\\over dx\\)(y) = \\(1\\over 2\\sqrt{x}\\)(\\(-cosec^2\\sqrt{x}\\))<\/p>\n
Hence, \\(d\\over dx\\)(\\(cot\\sqrt{x}\\)) = -\\(1\\over 2\\sqrt{x}\\)\\(cosec^2\\sqrt{x}\\)<\/p>\n
\nRelated Questions<\/h3>\n
What is the Differentiation of cot inverse x ?<\/a><\/p>\n