How do you differentiate #f(x)=(e^x+sinx)(cot^2x+1)# using the product rule?

1 Answer
Feb 24, 2018

Answer:

#(e^x+sinx)(-2cotxcsc^2x)+(cot^2x+1)(e^x+cosx)#

is the function obtained after differentiating using the product rule for the function described by

#f(x)=(e^x+sinx)(cot^2x+1)#

Explanation:

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)#