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