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Question
Problem Zero: Read the section of Chain Rule and make your own summary of the material.
Step-by-Step Solution
VerifiedThe section explains the chain rule, the gradient, and the directional derivatives.
If $$z = f(x_{1}, x_{2},...,x_{n})$$ and $$x_{i}= u_{i}(t_{1}), t_{2} ,...,t_{m})$$ for $$1 ≤ i ≤ n$$, for all values of $$t_{1}), t_{2} ,...,t_{m}$$ at which each $$u_{i}$$ is differentiable, and if $$f$$ is differentiable at $$u_{1}(t_{1}), t_{2} ,...,t_{m}), u_{2}(t_{1}), t_{2} ,...,t_{m}) ... , u_{3}(t_{1}), t_{2} ,...,t_{m})$$, then
$$\frac{\partial z }{\partial t_{j}}=\frac{\partial z }{\partial x_{1}}\frac{\partial x_{1} }{\partial t_{j}}+\frac{\partial z }{\partial x_{2}}\frac{\partial x_{2} }{\partial t_{j}}+,...,+\frac{\partial z }{\partial x_{n}}\frac{\partial x_{n} }{\partial t_{j}}$$, where, $$1\leq j\leq m$$
If $$w = f(x, y,z)$$ is a function of three variables, then the gradient of $$f$$ is the vector function defined by,
$$\bigtriangledown f(x,y,z)=\frac{\partial f}{\partial x} i+\frac{\partial f}{\partial y}j+\frac{\partial f}{\partial z}k={f_{x}(x,y,z), f_{y}(x,y,z), f_{z}(x,y,z)}$$
If $$f(x, y,z)$$ is a function of three variables, and $$(x_{0},y_{0},z_{0})$$
is a point in the domain of $$f$$ at which $$f$$ is differentiable, then for every unit vector $$u\epsilon R^{3}$$,
$$D_{u}f(x_{0},y_{0},z_{0})=\bigtriangledown f(x_{0},y_{0},z_{0})\cdot u$$