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eeek, i've totally forgotten how to do directional derivatives! Watch

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    two examples i've got here:

    f(x,y,z)= x^2 - y^2 - 2zx

    = 2(x-z)i - 2yj - 2xk

    and

    f(x,y,z)= ln(xy^2) + ln(y^2z) + ln(x^2z)

    = 3lnx + 4lny + 2lnz


    i know i have to derive but i've totally forgotten the rules- i don't know how to come to these conclusions! any help? thank you!
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    In general, \nabla f(x,y,z) = {\bf i} \frac{\partial f}{\partial x} + {\bf j} \frac{\partial f}{\partial y} + {\bf k} \frac{\partial f}{\partial z}.

    So, for the "i" component of the first one, \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (x^2-y^2-2zx) = 2x - 2z.

    The 2nd one isn't a directional derivative, it's just using normal log rules (log(ab) = log(a)+log(b), log(a^n) = n log a).
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    (Original post by DFranklin)
    In general, \nabla f(x,y,z) = {\bf i} \frac{\partial f}{\partial x} + {\bf j} \frac{\partial f}{\partial y} + {\bf k} \frac{\partial f}{\partial z}.

    So, for the "i" component of the first one, \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (x^2-y^2-2zx) = 2x - 2z.

    The 2nd one isn't a directional derivative, it's just using normal log rules (log(ab) = log(a)+log(b), log(a^n) = n log a).
    ahh thank you!! why didn't they do a directional derivative with the logs? is it not possible? and would i do it this way with any other kind of question i.e. sin, cos etc, or would they use the directional derivative formula? thank you!
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    Sure it's possible. I think what you posted is just the first line in a solution. They've simplified it using log rules. But then you just need to use the same method as for the first one.
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    oh my god, i've got it. thank you thank you!
 
 
 
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