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