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