Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie

maths help

Options
  • 31-05-2010 5:22pm
    #1
    Registered Users Posts: 82 ✭✭


    I'm having problems with differentiation help needed with the following and explanations explaining ur answer
    (i) if x=k(titha -sin titha)and y=k (1 - cos titha) find dy/dx
    (ii) e^-x squared sin x
    (iii)x^squarede^5x


Comments

  • Registered Users Posts: 319 ✭✭gemxpink


    (i) If x=k(θ - sinθ) and y=k (1 - cosθ), find dy/dx
    (ii) [LATEX]e^-^x^2 sin x[/LATEX]
    (iii) [LATEX]x^2 e^5^x[/LATEX]

    Just gonna leave that there.


  • Registered Users Posts: 167 ✭✭RexMundi


    Presuming that I read the questions right, these should be correct.

    (i) What you need to know is that dy/dx = dy/dt // dy/dx

    y = k(1-cost)
    .: = k-kcost

    .: dt/dt = ksint

    x= k(t-sint)
    .:= kt - ksint

    .: dx/dt = k - kcost

    but dy/dx = dy/dt // dx/dt = ksint/kt-kcost

    .: ANS = sint / t - cost

    (ii) Both this and the next one require the product rule.

    ie. if y = uv

    dy/dx = u(dv/dx) + v(du/dx)

    .: if y = e^-xx.sinx

    dy/dx = e^-xx.cosx - 2xe^-xx.sinx

    (iii) if y = xx.e^5x

    dy/dx = xx.5e^5x +e^5x.2x

    .: ANS = 5x^2.e^5x + 2x.e^5x


  • Registered Users Posts: 484 ✭✭brownacid


    Just use the product rule on all three cases


Advertisement