解
y'=[e^sin(x²+√x)]'
=e^sin(x²+√x)×[sin(x²+√x)]'
=e^sin(x²+√x)×cos(x²+√x)×(x²+√x)'
=cos(x²+√x)e^sin(x²+√x)×[2x+1/(2√x)]
=[2x+1/(2√x)]cos(x²+√x)e^sin(x²+√x)
∴dy={[2x+1/(2√x)]cos(x²+√x)e^sin(x²+√x)}dx