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Thread: Another partial derivative problem...

  1. #1
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    Another partial derivative problem...

    If f(x,y,z) = (x^3 + y^2 + z)^4 find each of the following:

    a) fx(x,y,z)
    b) fy(0,1,1)
    c) fzz(x,y,z)

    For a) I got 4(x^3 + y^2 + z)(3x^2)

    b) 4(x^3 + y^2 + z)(2y) which then comes out to 8 once you plug in everything

    Now c) is where I am lost. I have no clue how to differentiate fzz
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  2. #2
    knight.236
    Guest
    Differentiate twice with respect to z:

    1) Regard x and y as constants and use the chain rule:

    u = z + x^3 + y^2
    f(u)= u^4
    f'(u)*u'= 4(z + x^3 + y^2)^3 * 1

    f_z = 4(z + x^3 + y^2)^3

    2) Chain rule again:

    u = z + x^3 + y^2
    f(u)= 4(u^4)
    f'(u)*u'= 3*4(z + x^3 + y^2)^2 * 1

    f_zz = 12(z + x^3 + y^2)^2
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  3. #3
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    Hello, pakman!

    If $\displaystyle f(x,y,z) \:= \:(x^3 + y^2 + z)^4$, find each of the following:

    . . $\displaystyle (a)\; f_x(x,y,z)\qquad(b)\;f_y(0,1,1)\qquad(c)\; f_{zz}(x,y,z)$

    Now (c) is where I am lost. .I have no clue how to differentiate $\displaystyle f_{zz }$
    Just as $\displaystyle f_z$ means "differentiate with respect to $\displaystyle z$",
    . . $\displaystyle f_{zz}$ means "do it twice".


    We have: .$\displaystyle f \;=\;(x^3+y^2+z)^4$

    Then: .$\displaystyle f_z\;=\;4(x^3+y^2+z)^3\cdot1 \;=\;4(x^3+y^2+z)^3$

    Finally: .$\displaystyle f_{zz}\;=\;3\cdot4(x^3+y^2+z)^2\cdot1 \;=\;12(x^3+y^2+z)^2$

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