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Significance of derivative for functions of several variables

  1. Jun 28, 2012 #1
    For single variable functions derivative means slope of tangent what does it mean for functions of more than one variable.
    book says that a function is said to be differentiable if:
    f(x + Δx , y + Δy) - f(x , y) = AΔx + BΔy + ε'ψ(Δx , Δy) + εh(Δx , Δy)
    WHERE ε, ε' → 0 AS Δx , Δy → 0.
    PLEASE PROVIDE SOME ASSISTANCE ON IT. :confused: :cry: :confused:

  2. jcsd
  3. Jun 28, 2012 #2

    Simon Bridge

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    The derivative is the slope of the tangent to a "surface" in a particular direction. In the so f'(x=a) is usually the slope of the tangent to f at x=a in the x direction.
  4. Jun 28, 2012 #3


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    AAQIB IQBAL is talking about "the derivative" as if it were a single number and, for functions of several variables, that is not true.

    Simon Bridge is talking about the partial derivatives- imagine walking along a hillside (representing the surface z= f(x,y)) moving due east (the positive x-axis). The rate at which you go up or down is the partial derivative with respect to x. Moving northward (the positive y-axis) the rate at which you go up or down is the partial derivative with respect to y.

    The closest thing to "THE" derivative is the gradient, [itex]\nabla f[/itex], which is a vector pointing in the direction of fastest increase whose length is the rate of increase in that direction. Just as the "tangent line", having slope [itex]f'(x_0)[/itex], to a curve in the plane gives the "best" linear approximation to y= f(x) close to [itex]x= x_0[/itex], the "tangent plane, [itex]z= f_x(x_0,y_0)(x- x_0)+ f_y(x_0,y_0)(y- y_0)+ z_0[/itex] gives the best linear approximation to z= f(x,y) close to [itex](x_0, y_0)[/itex]. In terms of the gradient, that can be written [itex]z= \nabla f(x_0,y_0)\cdot <x- x_0, y- y_0>+ z_0[/itex] where that multiplication is the dot product of vectors.

    The formula your book gives is actually the equation of that tangent plane where the left side is "[itex]z- z_0[/itex]", [itex]\Delta x= x- x_0[/itex], [itex]\Delta y= y- y_0[/itex] and the last two terms are the "error" terms- the distance from the tangent plane to the actual surface.
  5. Jun 28, 2012 #4
    thank you, you have provided most satisfactory answer to my problem (especially "... imagine walking along ...").
    Now i need to know one thing that can we assume the coefficient of z in the equation of tangent plane is non zero without loss of generality.
    Actually while deriving the equation i need to divide throughout by coefficient of z, that is why i am asking this question.
  6. Jun 28, 2012 #5


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    IF you are given the surface as "z= f(x,y)" that is necessarily true. The most general form for a surface is "F(x,y,z)= constant", as, for example, the surface of a sphere, given by [itex]x^2+ y^2+ z^2= r^2[/itex]. In that case, the gradient of F, [itex]\nabla F[/itex], is normal to the surface at any point and the tangent plane to F(x,y,z)= C at [itex](x_0, y_0, z_0)[/itex] is [itex]\nabla\cdot<x- x_0, y- y_0, z- z_0>= 0[/itex]. In the case that z= f(x, y), we can write F(x, y, z)= f(x,y)- z= 0 so that [itex]\nabla F= <f_x, f_y, -1>[/itex] and the tangent plane is given by [itex]f_x(x- x_0)+ f_y(z- z_0)- (z- f(x_0, y_0))= 0[/itex]. So the coefficient of z, in the case that z= f(x,y), is non-zero.
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