{"id":5529,"date":"2021-09-16T20:30:35","date_gmt":"2021-09-16T15:00:35","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5529"},"modified":"2021-11-22T17:34:52","modified_gmt":"2021-11-22T12:04:52","slug":"differentiation-of-tan-inverse-x","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/differentiation-of-tan-inverse-x\/","title":{"rendered":"Differentiation of tan inverse x"},"content":{"rendered":"
Here you will learn differentiation of tan inverse x or arctanx x by using chain rule.<\/p>\n
Let’s begin –<\/p>\n
\nThe differentiation of \\(tan^{-1}x\\) with respect to x is \\(1\\over {1 + x^2}\\).<\/p>\n
i.e. \\(d\\over dx\\) \\(tan^{-1}x\\) = \\(1\\over {1 + x^2}\\).<\/p>\n<\/blockquote>\n
Proof using chain rule :<\/h2>\n
\nLet y = \\(tan^{-1}x\\). Then,<\/p>\n
\\(tan(tan^{-1}x)\\) = x<\/p>\n
\\(\\implies\\) tan y = x<\/p>\n
Differentiating both sides with respect to x, we get<\/p>\n
\\(d\\over dx\\)(tan y) = \\(d\\over dx\\)(x)<\/p>\n
\\(d\\over dx\\) (tan y) = 1<\/p>\n
By chain rule,<\/p>\n
\\(sec^2 y\\) \\(dy\\over dx\\) = 1<\/p>\n
\\(dy\\over dx\\) = \\(1\\over sec^2 y\\)<\/p>\n
[ \\(\\because\\) 1 + \\(tan^2 y\\) = \\(sec^2 y\\)<\/p>\n
\\(dy\\over dx\\) = \\(1\\over {1 + tan^2 y}\\)<\/p>\n
\\(\\implies\\) \\(dy\\over dx\\) = \\(1\\over {1 + x^2}\\)<\/p>\n
\\(\\implies\\) \\(d\\over dx\\) \\(tan^{-1}x\\) = \\(1\\over {1 + x^2}\\)\u00a0<\/p>\n
Hence, the differentiation of \\(tan^{-1}x\\) with respect to x is \\(1\\over {1 + x^2}\\).<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : What is the differentiation of \\(tan^{-1} x^2\\) with respect to x ?<\/p>\n
Solution<\/strong><\/span> : Let y = \\(tan^{-1} x^2\\)<\/p>\n
Differentiating both sides with respect to x and using chain rule, we get<\/p>\n
\\(dy\\over dx\\) = \\(d\\over dx\\) (\\(tan^{-1} x^2\\))<\/p>\n
\\(dy\\over dx\\) = \\(1\\over {1 + x^4}\\).(2x) = \\(2x\\over {1 + x^4}\\)<\/p>\n
Hence, \\(d\\over dx\\) (\\(tan^{-1} x^2\\)) = \\(2x\\over {1 + x^4}\\)<\/p>\n
Example<\/strong><\/span> : What is the differentiation of 2x + \\(tan^{-1} x\\) with respect to x ?<\/p>\n
Solution<\/strong><\/span> : Let y = 2x + \\(tan^{-1} x\\)<\/p>\n
Differentiating both sides with respect to x, we get<\/p>\n
\\(dy\\over dx\\) = \\(d\\over dx\\) (2x) + \\(d\\over dx\\) (\\(tan^{-1} x\\))<\/p>\n
\\(dy\\over dx\\) = 2 + \\(1\\over {1 + x^2}\\)<\/p>\n
Hence, \\(d\\over dx\\) (2x + \\(tan^{-1} x\\)) = 2 + \\(1\\over {1 + x^2}\\)<\/p>\n
\nRelated Questions<\/h3>\n
What is the Differentiation of tanx ?<\/a><\/p>\n
What is the Integration of Tan Inverse x ?<\/a><\/p>\n
What is the Differentiation of cos inverse x ?<\/a><\/p>\n\n\n