{"id":5525,"date":"2021-09-16T20:23:07","date_gmt":"2021-09-16T14:53:07","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5525"},"modified":"2021-11-22T17:44:23","modified_gmt":"2021-11-22T12:14:23","slug":"differentiation-of-sin-inverse-x","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/differentiation-of-sin-inverse-x\/","title":{"rendered":"Differentiation of sin inverse x"},"content":{"rendered":"
Here you will learn differentiation of sin inverse x or arcsin x by using chain rule.<\/p>\n
Let’s begin –<\/p>\n
\nIf x \\(\\in\\) (-1, 1) , then the differentiation of \\(sin^{-1}x\\) with respect to x is \\(1\\over \\sqrt{1 – x^2}\\).<\/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
Proof using chain rule :<\/h2>\n
\nLet y = \\(sin^{-1}x\\). Then,<\/p>\n
\\(sin(sin^{-1}x)\\) = x \\(\\implies\\) sin y = x<\/p>\n
Differentiating both sides with respect to x, we get<\/p>\n
1 = \\(d\\over dx\\) (sin y)<\/p>\n
By chain rule,<\/p>\n
1 = \\(d\\over dx\\) (sin y) \\(\\times\\) \\(dy\\over dx\\)<\/p>\n
1 = cos y \\(dy\\over dx\\)<\/p>\n
\\(dy\\over dx\\) = \\(1\\over cos y\\)<\/p>\n
\\(dy\\over dx\\) = \\(1\\over \\sqrt{1 – sin^2 y}\\)<\/p>\n
\\(\\implies\\) \\(dy\\over dx\\) = \\(1\\over \\sqrt{1 – x^2}\\)<\/p>\n
\\(\\implies\\) \\(d\\over dx\\) \\(sin^{-1}x\\) = \\(1\\over \\sqrt{1 – x^2}\\)\u00a0<\/p>\n
Hence, the differentiation of \\(sin^{-1}x\\) with respect to x is \\(1\\over \\sqrt{1 – x^2}\\).<\/p>\n<\/blockquote>\n
Example<\/span><\/strong> : What is the differentiation of \\(sin^{-1} x^3\\) with respect to x ?<\/p>\n
Solution<\/span><\/strong> : Let y = \\(sin^{-1} x^3\\)<\/p>\n
Differentiating both sides with respect to x and using chain rule, we get<\/p>\n
\\(dy\\over dx\\) = \\(d\\over dx\\) (\\(sin^{-1} x^3\\))<\/p>\n
\\(dy\\over dx\\) = \\(1\\over \\sqrt{1 – x^6}\\).\\(3x^2\\) = \\(3x^2\\over \\sqrt{1 – x^6}\\)<\/p>\n
Hence, \\(d\\over dx\\) (\\(sin^{-1} x^3\\)) = \\(3x^2\\over \\sqrt{1 – x^6}\\)<\/p>\n
Example<\/span><\/strong> : What is the differentiation of \\(x^2\\) + \\(sin^{-1} x^5\\) with respect to x ?<\/p>\n
Solution<\/span><\/strong> : Let y = \\(x^2\\) + \\(sin^{-1} x^5\\)<\/p>\n
Differentiating both sides with respect to x and using chain rule, we get<\/p>\n
\\(dy\\over dx\\) = \\(d\\over dx\\) (\\(x^2\\)) + \\(d\\over dx\\) (\\(sin^{-1} x^5\\))<\/p>\n
\\(dy\\over dx\\) = 2x + \\(1\\over \\sqrt{1 – x^{10}}\\).\\(5x^4\\)<\/p>\n
Hence, \\(d\\over dx\\) (\\(x^2\\) + \\(sin^{-1} x^5\\))= 2x + \\(5x^4\\over \\sqrt{1 – x^{10}}\\)<\/p>\n
\nRelated Questions<\/h3>\n
What is the Differentiation of sin x ?<\/a><\/p>\n
What is the Integration of Sin Inverse x ?<\/a><\/p>\n
What is the Differentiation of sec inverse x ?<\/a><\/p>\n\n\n