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Mathematics
Calculus
Derive the formula for gradient using chain rule
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[QUOTE="Happiness, post: 5503259, member: 549579"] The following website calculates the gradient as follows: Differentiate ##g## wrt ##x## while holding ##z## constant. [ATTACH=full]102281[/ATTACH] Then we will have ##\frac{\partial y}{\partial x}=-\frac{(\frac{\partial g}{\partial x})}{(\frac{\partial g}{\partial y})}##. Why ain't ##\frac{\partial g}{\partial x}## and ##\frac{\partial g}{\partial y}## zero in this case? In what situations must the two partials on the RHS of the chain rule have the same variables held constant? Like you mentioned for post #1, ##(\frac{\partial y}{\partial g})_{gz}(\frac{\partial g}{\partial x})_{gz}##. But why is it not required in the attachment above, where ##(\frac{\partial g}{\partial y})_{xz}(\frac{\partial y}{\partial x})_{gz}##? Source: http://www.sjsu.edu/faculty/watkins/envelopetheo.htm [/QUOTE]
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Mathematics
Calculus
Derive the formula for gradient using chain rule
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