Chain Rule (1 of 3: Introducing a substitution)
Chain Rule (1 of 3: Introducing a substitution)

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Let f: R^n → R^m and g: R^m → R^p be functions.
Write down what it means for f to be differentiable at an arbitrary x ∈ R^n and for g to be differentiable at an arbitrary y ∈ R^m.
(b) Let h = g∘f. State the chain rule for h'(x), the matrix of the derivative of h, at point x, in terms of the matrices of the derivatives of g and of f.
(c) Write down a sufficient condition on the partial derivatives of g and of f such that the entries of the derivative matrices above are given by the partial derivatives.
Let h = g∘f, as in (6):
Under the condition in (c), use the partial derivatives to write down a summation expression for Dshk (x) in terms of the partial derivatives of g and of f. Be sure to label the summation indices appropriately.
In the case that m = p = n and f and g are given by f_i(x) = x_i and g_k(y) = ky_r + 1, for i, k, use the above to write down expressions for Dihi (x) and Dhi (x,0,0).

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02:32

Given g(x, y) = (x^2 + 1, y^2) and f(u, v) = (u + v, u, v^2), calculate the (matrix) derivative of f•g at the point (11) using the chain rule. (Partial derivatives in polar coordinates) Given the function f(x, y), make the substitution x = r cos(θ), y = r sin(θ) and find, using the chain rule, a formula for the derivatives ∂f/∂θ and ∂f/∂r at the point (r, θ) (Note: the result must be in terms of the partial derivatives ∂f/∂x and ∂f/∂y evaluated at (r cos θ, r sin θ) and functions of r and θ).

01:31

If we write $g(x)$ for $f^{-1}(x)$, Equation (1) can be written as$$g^{\prime}(f(a))=\frac{1}{f^{\prime}(a)^{\prime}} \quad \text { or } \quad g^{\prime}(f(a)) \cdot f^{\prime}(a)=1$$If we then write $x$ for $a,$ we get$$g^{\prime}(f(x)) \cdot f^{\prime}(x)=1$$The latter equation may remind you of the Chain Rule, and indeed mere is a connection.

Assume that $f$ and $g$ are differentiable functions that are inverses of one another, so that $(g \circ f)(x)=x .$ Differentiate both sides of this equation with respect to $x$, using the Chain Rule to $=$ xpress $(g \circ f)^{\prime}(x)$ as a product of derivatives of $g$ and $f .$ What do Jou find? (This is not a proof of Theorem 3 because we assume here the theorem’s conclusion that $g=f^{-1}$ is differentiable.)

09:48

Show that $\nabla(u v)=v \nabla u+u \nabla v$, where $u$ and $v$ are differentiable scalar functions of $x, y$, and $z$.(a) Show that a necessary and sufficient condition that $u(x, y, z)$ and $v(x, y, z)$ are related by some function $f(u, v)=0$ is that $(\nabla u) \times(\nabla v)=0$.(b) If $u=u(x, y)$ and $v=v(x, y)$, show that the condition $(\nabla u) \times(\nabla v)=0$ leads to the two-dimensional Jacobian$$J\left(\frac{u, v}{x, y}\right)=\left|\begin{array}{ll}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{array}\right|=0 .$$The functions $u$ and $v$ are assumed differentiable.

03:04

Define the function F of two variables by F(x,y) = x*f(x,y,g(x,y)), where f and g are differentiable functions. Find an expression for the partial derivative of F with respect to x in terms of the partial derivatives of f and g.

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You are watching: SOLVED: Let f: R^n → R^m and g: R^m → R^p be functions. Write down what it means for f to be differentiable at an arbitrary x ∈ R^n and for g to be differentiable at an arbitrary y ∈ R^m. (b) . Info created by Bút Chì Xanh selection and synthesis along with other related topics.