Differentiation of a product: e^t ln(t) cos(t)

[Anthonyk2013]

Wondering if correct could this be simplified more..

\(\displaystyle 8.\, e^t\, \ln(t)\, \cos(t)\)

\(\displaystyle u\, =\, e^t\, \ln(t)\). . .\(\displaystyle \dfrac{du}{dt}\, =\, e^t\, \dfrac{1}{t}\)

\(\displaystyle v\, =\, \cos(t)\). . .\(\displaystyle \dfrac{dv}{dt}\, =\, -\sin(t)\)

\(\displaystyle \dfrac{dy}{dt}\, =\, \bigg(\, e^t\, \ln(t)\, \bigg)\, \bigg(\, -\sin(t)\, \bigg)\, +\, \bigg(\, \cos(t)\, \bigg)\, \bigg(\, e^t\, \dfrac{1}{t}\, \bigg)\)

 

[Deleted member 4993]

Wondering if correct could this be simplified more..

\(\displaystyle 8.\, e^t\, \ln(t)\, \cos(t)\)

\(\displaystyle u\, =\, e^t\, \ln(t)\). . .\(\displaystyle \dfrac{du}{dt}\, =\, e^t\, \dfrac{1}{t}\)

\(\displaystyle v\, =\, \cos(t)\). . .\(\displaystyle \dfrac{dv}{dt}\, =\, -\sin(t)\)

\(\displaystyle \dfrac{dy}{dt}\, =\, \bigg(\, e^t\, \ln(t)\, \bigg)\, \bigg(\, -\sin(t)\, \bigg)\, +\, \bigg(\, \cos(t)\, \bigg)\, \bigg(\, e^t\, \dfrac{1}{t}\, \bigg)\)

Incorrect - there are three (not two) functions of 't' [ e^t ; ln(t) & cos(t)].

Extend the product rule to three functions.

 

[Anthonyk2013]

\(\displaystyle 8.\, e^t\, \ln(t)\, \cos(t)\)

\(\displaystyle u\, =\, e^t\, \ln(t)\). . .\(\displaystyle \dfrac{du}{dt}\, =\, e^t\, \dfrac{1}{t}\)

\(\displaystyle v\, =\, \cos(t)\). . .\(\displaystyle \dfrac{dv}{dt}\, =\, -\sin(t)\)

\(\displaystyle \dfrac{dy}{dt}\, =\, \bigg(\, e^t\, \ln(t)\, \bigg)\, \bigg(\, -\sin(t)\, \bigg)\, +\, \bigg(\, \cos(t)\, \bigg)\, \bigg(\, e^t\, \dfrac{1}{t}\, \bigg)\)

Incorrect - there are three (not two) functions of 't' [ e^t ; ln(t) & cos(t)].

Extend the product rule to three functions.

so I should ha e^t separate from lnt ?

 

[Deleted member 4993]

so I should ha e^t separate from lnt ?

yes