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Derivatives of Trigonometric Functions

  1. Oct 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Find d^2y/dx^2.

    y = x cos x

    3. The attempt at a solution

    I've been doing derivatives recently and now got into doing them with trig functions.

    I thought it was,

    y = x cos x = -xsinx = -xcosx

    but that is the derivative of the derivative.

    The problem asks for d^2/dx^2.

    Can someone explain what to do in this situation?

    What is the difference of dx/dy, d^2y/dx^2, dy^2/d^2x, etc....

    Please and thank you. :smile:
     
  2. jcsd
  3. Oct 8, 2007 #2

    cristo

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    You need to use the product rule to differentiate x.cosx; i.e. d/dx(uv)=v(du/dx)+u(dv/dx). To find the second derivative you simply differentiate the result of d/dx(x.cosx).
     
  4. Oct 8, 2007 #3

    CompuChip

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    First find the first derivative. You get two terms (product rule!).

    dx/dy is the derivative of x with respect to y. Writing this down sort of implies that x is a function of y, like [itex]x(y) = 3 \sqrt{1 + y^2} / y[/itex], though it's a bit against conventions to call the function x and the variable y.

    d^2 y / dx^2 is the second derivative of y with respect to x, of which y is a function. It can be viewed as shorthand for
    [tex] \frac{d^2 y}{dx^2} = \frac{d\left( \frac{ dy }{ dx } \right) }{ dx } [/tex]
    Of course this is possible, since dy/dx is again a function of x, for example:
    [tex]y(x) = x^2, \frac{dy}{dx} = 2 x, \frac{d^2 y}{dx} = \frac{d (2x) }{dx} = 2.[/tex]

    Oh, and dy^2/dx^2 doesn't really mean anything, it's usually an error if someone writes that down. (Though it is possible to make sense of it, by viewing x^2 as a variable, e.g. [itex]y = 2 x^2[/itex] s.t. [itex]y^2 = 4 (x^2)^2 = 4 x^4[/itex]. Then [itex]\frac{ d y^2 }{ d x^2 } = 8 x^2[/itex] as you can see by temporarily writing [itex]z = x^2[/itex] everywhere, deriving, and writing it back in x^2).
     
    Last edited: Oct 8, 2007
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