Homework Help

What is the derivative dx/dt given that x*t + x^2*sin t = 3*x^3*t^2 + 2t

user profile pic

needyourhelp | Student, Kindergarten | (Level 1) eNoter

Posted September 3, 2012 at 1:08 PM via web

dislike 1 like

What is the derivative dx/dt given that x*t + x^2*sin t = 3*x^3*t^2 + 2t

1 Answer | Add Yours

user profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted September 3, 2012 at 1:17 PM (Answer #1)

dislike 1 like

The derivative `dx/dt` has to be determined given that `x*t + x^2*sin t = 3*x^3*t^2 + 2t` . This can be done using implicit differentiation.

`(d(x*t + x^2*sin t))/dt = (d(3*x^3*t^2 + 2t))/dt`

=> `(dx/dt)*t + x + 2*x*(dx/dt)*sin t + x^2*cos t = 9*x^2*(dx/dt)*t^2 + 6*x^3*t + 2`

=> `(dx/dt)(t + 2x*sin t - 9*x^2*t^2) = 6*x^3*t - x - x^2*cos t + 2`

=> `(dx/dt) = (6*x^3*t - x - x^2*cos t + 2)/(t + 2x*sin t - 9*x^2*t^2)`

The derivative `dx/dt = (6*x^3*t + 2 - x - x^2*cos t + 2)/(t + 2x*sin t - 9*x^2*t^2)`

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes