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