Solution :
Let y = \(e^{sinx}\). Putting u = sinx , we get
y = \(e^u\) and u = sinx
\(\therefore\)ย \(dy\over du\) = \(e^u\) and \(du\over dx\) = cosx
Now, \(dy\over dx\) = \(dy\over du\) \(\times\) \(du\over dx\)
\(\implies\) \(dy\over dx\) = \(e^u\)cosx = \(e^{sinx}\)cosx
Hence, the differentiation of \(e^{sinx}\) with respect to x is \(e^{sinx}\)cosx.
Questions for Practice
What is the differentiation of cosx sinx ?
What is the differentiation of sin square x or \(sin^2x\) ?
What is the differentiation of 1/sinx ?
