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Natalie Goodwin

Feb 20, 2025

If z = f(x, y), where f is differentiable, and the following applies, find dz/dt when t = 1.?

x = g(t) y = h(t) g(1) = -5 h(1) = 1 g’ (1) = -6 h’ (1) = 1 fx(-5, 1) = 1 fy(-5, 1) = -1
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Miss Alice Doyle

Feb 20, 2025

laughing is correct up til"dz/dt = -6 1 + 1 -1dz/dt = -6 – 2dz/dt = -8" It’s easy to fix though (just 1 * -1 = -1, not -2):dz/dt = -6 1 + 1 -1dz/dt = -6 – 1dz/dt = -7... Show More
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Miss Alice Doyle

Feb 20, 2025

If z = f(x,y)(dz/dt) = (dx/dt) (dz/dx) + (dy/dt) (dz/dy) Where dz/dx and dz/dy are partial derivatives. I am not sure what fx(-5,1) = 1 and fy(-5,1) = -1 mean, but I guess they represent these partial derivatives. Summarizing the information until now:g(1) = -5 –> x = -5 in this special case for t = 1h(1) = 1 –> y = 1 in this special case for t = 1 g'(1) = -6h'(1) = 1for those specific values of x and y, the partial derivatives are:dz/dx = 1dz/dy = -1Filling in gives:(dz/dt) = (dx/dt) (dz/dx) + (dy/dt) (dz/dy)dz/dt = -6 1 + 1 -1dz/dt = -6 – 2 dz/dt = -8Edit: I have to admit, that’s a bit stupid :). Thanks for correcting me!... Show More
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