Question

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

Answer 1

`(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)`