Solution :
Let y = cosx.sinx
By using product rule in differentiation,
\(dy\over dx\) = sinx(-sinx) + cosx.cosx
\(dy\over dx\) = \(cos^2x โ sin^2x\) = cos 2x
Hence, the differentiation of cosx.sinx with respect to x is cos 2x.
Questions for Practice
What is the differentiation of \(e^{sinx}\) ?
What is the differentiation of sin square x or \(sin^2x\) ?
What is the differentiation of 1/sinx ?
