Find a unit vector in the direction in which f decreases most rapidly at (1,1,-1)?

f(x,y,z) = ye^(y - x) - z²

Remember when finding the partial derivative with respect to y to use the product rule:

∇f = <-ye^(y - x), ye^(y - x) + e^(y - x), 2z>

= <-ye^(y - x), e^(y - x)(y + 1), 2z>

At the point (1, 1, -1), we have:

∇f(1, 1, -1) = <-1, 2, -2>

And the magnitude is given by:

|∇f(1, 1, -1)| = √(1 + 4 + 4) = 3

The unit vector in which f is decreasing most rapidly is given by:

u = ∇f(1, 1, -1) / |∇f(1, 1, -1)|

= <-1, 2, -2> / 3

= <-1/3, 2/3, -2/3>

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