Given:
f(x)=(e^x+sinx)(cot^2x+1)
Let
y=f(x)
u=(e^x+sinx)
v=(cot^2x+1)
By the product rule,
d/(dx)(uv)=u(dv)/(dx)+v(du)/(dx)
Here,
u=(e^x+sinx)
Differentiating wrt x
(du)/(dx)=d/(dx)(u)
d/(dx)(u)=d/(dx)(e^x+sinx)
By the sum rule
d/(dx)(e^x+sinx)=d/(dx)(e^x)+d/(dx)(sinx)
d/(dx)(e^x)=e^x
d/(dx)(esinx)=cosx
d/(dx)(e^x+sinx)=e^x+cosx
d/(dx)(u)=e^x+cosx
(du)/(dx)=e^x+cosx
v=(cot^2x+1)
Differentiating wrt x
(dv)/(dx)=d/(dx)(v)
d/(dx)(v)=d/(dx)(cot^2x+1)
By the sum rule
d/(dx)(cot^2x+1)=d/(dx)(cot^2x)+d/(dx)(1)
d/(dx)(cot^2x)=2cotx(-csc^2x)
d/(dx)(cot^2x)=-2cotxcsc^2x
d/(dx)(1)=0
d/(dx)(cot^2x+1)=-2cotxcsc^2x+0
d/(dx)(v)=-2cotxcsc^2x
(dv)/(dx)=-2cotxcsc^2x
d/(dx)(uv)=u(dv)/(dx)+v(du)/(dx)
u=(e^x+sinx)
v=(cot^2x+1)
(du)/(dx)=e^x+cosx
(dv)/(dx)=-2cotxcsc^2x
d/(dx)((e^x+sinx)(cot^2x+1))=(e^x+sinx)(-2cotxcsc^2x)+(cot^2x+1)(e^x+cosx)
