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