Directional derivative tangent to a curve

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SUMMARY

The discussion focuses on calculating the directional derivative of the implicitly defined function z, given by the equation xz² - yz + cos(xy) = 2. The specific task is to find this derivative in the direction of the tangent to the curve y = x² + 2x - 1 at the point (0, 1), specifically in the direction of decreasing x. The solution involves using the negative gradient of z and the unit vector of the tangent, which is derived from the derivative of the curve, dy/dx = 2x + 2, evaluated at the point of interest.

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Homework Statement



Given xz^2-yz+cos(xy)=2 which defines z implicitly in terms of x and y, find the directional derivative of z in the direction of the tangent to the curve y=x^2+2x-1 at the point (0,1) in the direction of decreasing x

Homework Equations





The Attempt at a Solution


I'm fairly positive I just take the negative of the gradient of z multiplied by the unit vector of the tangent of the curve, all at the point. I have the gradient of z but what is the unit vector of the tangent? I think the tangent would be 2x+2 but I know it has to be divided by something to be a unit vector. This is where my notes become inconsistent and I'm not sure what to do
 
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If you draw in the tangent vector to the curve, you should be able to see that \tan \theta = dy/dx, where \theta is the angle the vector makes with the horizontal axis.
 

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