{"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

Differentiation of tan inverse x or \\(tan^{-1}x\\) :<\/h2>\n
\n

The 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
\n

Let 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


\n

Related 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

\n
Next – Differentiation of cot inverse x<\/a><\/div>\n<\/div>\n\n\n\n
\n
Previous – Differentiation of cos inverse x<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Here you will learn differentiation of tan inverse x or arctanx x by using chain rule. Let’s begin – Differentiation of tan inverse x or \\(tan^{-1}x\\) : The differentiation of \\(tan^{-1}x\\) with respect to x is \\(1\\over {1 + x^2}\\). i.e. \\(d\\over dx\\) \\(tan^{-1}x\\) = \\(1\\over {1 + x^2}\\). Proof using chain rule : Let …<\/p>\n

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