how to solve this calculus problem?

the slope of the line normal to the curve y=xe^sinx at x=2.438

2 Answers

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  • 8 years ago

    firstly, slope=gradient=dy/dx

    from y=xe^sinx

    dy/dy = x.cosx.e^sinx + e^sinx (using product rule here)

    At x=2.438

    dy/dx = (2.438).cos(2.438).e^sin(2.438) + e^sin(2.438)

    dy/dx = 3.585

    But this is the gradient of the curve at x=2.438

    The gradient of the normal to the curve = -1/3.585

  • 8 years ago

    Download Graph 4.4 from www.padowan.dk for free.

    On "Function I Insert function", type "x*exp(sin(x))", then "OK".

    Click on the 7th icon, type "2.438", then "OK".

    The normal line is 1.6405*x + 8.6555; the slope is 1.6405, answer!

    Source(s): Graph 4.4
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