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

Question

Isabella Ackerman · Accepted 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)# Given: #f(x)=(e^x+sinx)(cot^2x+1)# Let #y=f(x)# #u=(e^x+sinx)# #v=(cot^2x+1)# In accordance with the product rule, #d/(dx)(uv)=u(dv)/(dx)+v(du)/(dx)# Here, #u=(e^x+sinx)# Differing with respect to x #(du)/(dx)=d/(dx)(u)# #d/(dx)(u)=d/(dx)(e^x+sinx)# According to 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)# Differing with respect to x #(dv)/(dx)=d/(dx)(v)# #d/(dx)(v)=d/(dx)(cot^2x+1)# According to 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)#

Ezekiel Alexander · Answer

undefined To differentiate the function $ f(x) = (e^x + \sin(x))( \cot^2(x) + 1) $ using the product rule, follow these steps:

1. Let $ u(x) = e^x + \sin(x) $ and $ v(x) = \cot^2(x) + 1 $.
2. Find the derivatives of $ u(x) $ and $ v(x) $, denoted as $ u'(x) $ and $ v'(x) $, respectively.
3. Apply the product rule: $ f'(x) = u'(x)v(x) + u(x)v'(x) $.
4. Substitute the expressions for $ u'(x) $, $ v'(x) $, $ u(x) $, and $ v(x) $ into the product rule formula.
5. Simplify the expression to obtain the derivative $ f'(x) $.