{"id":5129,"date":"2021-09-08T20:34:50","date_gmt":"2021-09-08T15:04:50","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5129"},"modified":"2021-11-22T21:58:39","modified_gmt":"2021-11-22T16:28:39","slug":"differentiation-of-inverse-trigonometric-functions","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/differentiation-of-inverse-trigonometric-functions\/","title":{"rendered":"Differentiation of Inverse Trigonometric Functions"},"content":{"rendered":"
Here you will learn what is the differentiation of inverse trigonometric functions with examples.<\/p>\n
Let’s begin –<\/p>\n
(i)<\/strong> If x \\(\\in\\) (-1, 1), then the differentiation of \\(sin^{-1}x\\)<\/strong> or arcsinx<\/strong> with respect to x is \\(1\\over \\sqrt{1-x^2}\\)<\/strong>.<\/p>\n i.e. \\(d\\over dx\\) \\(sin^{-1}x\\) = \\(1\\over \\sqrt{1-x^2}\\) , for x \\(\\in\\) (-1, 1).<\/p>\n<\/blockquote>\n (ii)<\/strong> If x \\(\\in\\) (-1, 1), then the differentiation of \\(cos^{-1}x\\)<\/strong> or arccosx<\/strong> with respect to x is \\(-1\\over \\sqrt{1-x^2}\\)<\/strong>.<\/p>\n i.e. \\(d\\over dx\\) \\(cos^{-1}x\\) = \\(-1\\over \\sqrt{1-x^2}\\) , for x \\(\\in\\) (-1, 1).<\/p>\n<\/blockquote>\n (iii)<\/strong> The differentiation of \\(tan^{-1}x\\)<\/strong> or arctanx<\/strong> with respect to x is \\(1\\over {1+x^2}\\)<\/strong>.<\/p>\n i.e. \\(d\\over dx\\) \\(tan^{-1}x\\) = \\(1\\over {1+x^2}\\).<\/p>\n<\/blockquote>\n (iv)<\/strong> The differentiation of \\(cot^{-1}x\\)<\/strong> or arccotx<\/strong> with respect to x is \\(-1\\over {1+x^2}\\)<\/strong>.<\/p>\n i.e. \\(d\\over dx\\) \\(cot^{-1}x\\) = \\(-1\\over {1+x^2}\\).<\/p>\n<\/blockquote>\n (v)<\/strong> If x \\(\\in\\) R – [-1, 1], then the differentiation of \\(sec^{-1}x\\)<\/strong> or arcsecx<\/strong> with respect to x is \\(1\\over |x|\\sqrt{x^2-1}\\)<\/strong>.<\/p>\n i.e. \\(d\\over dx\\) \\(sec^{-1}x\\) = \\(1\\over |x|\\sqrt{x^2-1}\\) , x \\(\\in\\) R – [-1, 1]<\/p>\n<\/blockquote>\n (vi)<\/strong> If x \\(\\in\\) R – [-1, 1], then the differentiation of \\(cosec^{-1}x\\)<\/strong> or arccosecx<\/strong> with respect to x is \\(-1\\over |x|\\sqrt{x^2-1}\\)<\/strong>.<\/p>\n i.e. \\(d\\over dx\\) \\(cosec^{-1}x\\) = \\(-1\\over |x|\\sqrt{x^2-1}\\) , x \\(\\in\\) R – [-1, 1]<\/p>\n<\/blockquote>\n Example 1<\/span><\/strong> : find the differentiation of \\(sin^{-1}5x\\).<\/p>\n Solution<\/span><\/strong> : Let y = \\(sin^{-1}5x\\)<\/p>\n Now, \\(dy\\over dx\\) = \\(1\\over \\sqrt{1-(5x)^2}\\).5<\/p>\n = \\(5\\over \\sqrt{1-25x^2}\\)<\/p>\n Example 2<\/span><\/strong> : find the differentiation of \\(cos^{-1}5x\\).<\/p>\n Solution<\/span><\/strong> : Let y = \\(cos^{-1}5x\\)<\/p>\n Now, \\(dy\\over dx\\) = \\(-1\\over \\sqrt{1-(5x)^2}\\).5<\/p>\n = \\(-5\\over \\sqrt{1-25x^2}\\)<\/p>\n Example 3<\/span><\/strong> : find the differentiation of \\(tan^{-1}5x\\).<\/p>\n Solution<\/span><\/strong> : Let y = \\(tan^{-1}5x\\)<\/p>\n Now, \\(dy\\over dx\\) = \\(1\\over {1+(5x)^2}\\).5<\/p>\n = \\(5\\over {1+25x^2}\\)<\/p>\n Example 4<\/span><\/strong> : find the differentiation of \\(sec^{-1}5x\\).<\/p>\n Solution<\/span><\/strong> : Let y = \\(sec^{-1}5x\\)<\/p>\n Now, \\(dy\\over dx\\) = \\(1\\over |5x|\\sqrt{(5x)^2 – 1}\\).5<\/p>\n = \\(5\\over |5x|\\sqrt{25x^2-1}\\) = \\(1\\over x\\sqrt{25x^2-1}\\)<\/p>\n Example 5<\/span><\/strong> : find the differentiation of \\(cosec^{-1}5x\\).<\/p>\n Solution<\/span><\/strong> : Let y = \\(cosec^{-1}5x\\)<\/p>\n Now, \\(dy\\over dx\\) = \\(-1\\over |5x|\\sqrt{1-(5x)^2}\\).5<\/p>\n = \\(-5\\over |5x|\\sqrt{25x^2-1}\\) = \\(-1\\over x\\sqrt{25x^2-1}\\)<\/p>\n\n\n\n
\n
\n
\n
\n
\n