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  1. #1
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    partial der

    I am unsure where to begin with finding this partial derivative.
    f(x,y)=integral from y to x of cos(t^2)dt
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  2. #2
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    Quote Originally Posted by thehollow89 View Post
    I am unsure where to begin with finding this partial derivative.
    f(x,y)=integral from y to x of cos(t^2)dt
    f(x, y) = \int_y^x \cos (t^2) \, dt.

    Using the Fundamental Theorem of Calculus and noting that y is treated as a constant when you differentiate with respect to x:

    \frac{\partial f}{\partial x} = \cos(x^2).

    Now note that f(x, y) = \int_y^x \cos (t^2) \, dt = - \int_x^y \cos (t^2) \, dt and use the Fundamental Theorem of Calculus again (note that x is treated as a constant when you differentiate with respect to y):

    \frac{\partial f}{\partial y} = - \cos(y^2).
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  3. #3
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    Quote Originally Posted by thehollow89 View Post
    I am unsure where to begin with finding this partial derivative.
    f(x,y)=integral from y to x of cos(t^2)dt
    Depends on what partial derivative you want

    f(x,y) = \int_y^x \cos t^2\, dt = -\int_x^y \cos t^2 dt

    so using the 2nd fundemental theorem of Calculus

     \frac{\partial f }{\partial x } = \frac{\partial }{\partial x } \int_y^x \cos t^2 dt = \cos x^2

     \frac{\partial f }{\partial y } = - \frac{\partial }{\partial y } \int_x^y \cos t^2 dt = - \cos y^2
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