Question
Let and f be a continuously differentiable function of x and y, whose partial derivatives are also continuously differentiable. Show that
Answer :
Word Count : 530
We are given (x = e^r \cos\theta) and (y = e^r \sin\theta), and (f(x,y)) is continuously differentiable. We want to show: [ \frac{\partial^2 f}{\partial r^2} + \frac{\partial^2 f}{\partial \theta^2} = (x^2 + y^2) \left(\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}\right). ] --- First, compute the first derivatives of (x) and (y) with respect to (r) and (\theta): [ \frac{\partial x}{\partial r} = e^r \cos\theta = x, \quad \frac{\partial x}{\partial \theta} = -e^r \sin\theta = -y, ] [ \frac{\partial y}{\partial r} = e^r \sin\theta = y, \quad \frac{\partial y}{\partial ________ _______ ______ __________ ______.
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We are given (x = e^r \cos\theta) and (y = e^r \sin\theta), and (f(x,y)) is continuously differentiable. We want to show: [ \frac{\partial^2 f}{\partial r^2} + \frac{\partial^2 f}{\partial \theta^2} = (x^2 + y^2) \left(\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}\right). ] --- First, compute the first derivatives of (x) and (y) with respect to (r) and (\theta): [ \frac{\partial x}{\partial r} = e^r \cos\theta = x, \quad \frac{\partial x}{\partial \theta} = -e^r \sin\theta = -y, ] [ \frac{\partial y}{\partial r} = e^r \sin\theta = y, \quad \frac{\partial y}{\partial ________ _______ ______ __________ ______.
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