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it's not my homework, I just want to find out how to find a directional derivative of an implicit function. I know what is a directional derivative and how to find it when I have a function in normal form (I mean like like z=x^2+y....). $$ xz + yz^2 = 3xy + 3 $$ the point is: $$ P(1,−1) $$ and the direction (vector): $$ u = [1, 1] $$ Could you give me a formula for this? I know how to compute the derivatives of an implicit function as well.

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Implicit differentiation (officially, the Implicit Function Theorem if you rewrite your equation as $F(x,y,z)=0$) allows you to compute the gradient of $z=g(x,y)$, and then you use your usual dot product.

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  • $\begingroup$ $$ xz+yz^2-3xy-3=0 $$ and now how to create z=g(x,y)? I have no idea how, since there is $$ z $$ as well as $$ z^2 $$ in this equation. $\endgroup$ – TomDavies92 Jun 3 '13 at 15:57
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    $\begingroup$ You said you knew how to find the derivatives of an implicit function. Differentiate implicitly with respect to $x$ and solve for $\partial z/\partial x$, etc., just as you did in single-variable calculus.t $\endgroup$ – Ted Shifrin Jun 3 '13 at 16:28
  • $\begingroup$ I know how to find a derivative of an implicit function, I mean: $$ -\frac{\frac{∂F}{∂x}}{\frac{∂F}{∂y}} $$ so when I want to find a partial derivative e.g. with respect to x, I do: $$ \frac{∂z}{∂x} $$ and is it correct for an implicit function as well? $\endgroup$ – TomDavies92 Jun 3 '13 at 16:35
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    $\begingroup$ Your first fraction gives you $dy/dx$ if you have $F(x,y)=0$. If you have $F(x,y,z)=0$, then $$\frac{\partial z}{\partial x} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}\,.$$ If you just do basic "implicit differentiation" as a beginning calculus student would, this is exactly what you get! $\endgroup$ – Ted Shifrin Jun 3 '13 at 16:39
  • $\begingroup$ ok, so the partial derivative of this implicit function with respect to x is $$ -\frac{z-3y}{z^2-3x} $$ ?? $\endgroup$ – TomDavies92 Jun 3 '13 at 16:45
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LOL that guy wouldn't help you at all, its better if you think of it as like this,

(dz/dx) = -(Fx/Fz) ;this is the standard formula

that is a negative times, the derivative of f wrt x divided by derivative of f wrt to z.

so it would be,

(dz/dx) = -(z+3y)/(x+y*2z)

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    $\begingroup$ Welcome to Math.SE! Could you improve your answer somewhat? Right now it is pretty hard to read: What is Fx supposed to be, $F(x)$, $\partial_x F$, $F(x) x$? $\endgroup$ – Hrodelbert Oct 23 '15 at 8:30

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