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Answers without the blur. Sign up and see all textbooks for free! Q. 76

Expert-verified Found in: Page 946 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # $\mathrm{Let}f\left(x\right)\mathrm{be}\mathrm{a}\mathrm{differentiable}\mathrm{function}\mathrm{of}x,g\left(y\right)\mathrm{be}\mathrm{a}\mathrm{differentiable}\mathrm{function}\mathrm{of}y,\phantom{\rule{0ex}{0ex}}\mathrm{and}h\left(x,y\right)=f\left(x\right)g\left(y\right).\mathrm{Prove}\mathrm{that}\frac{{\partial }^{2}h}{\partial x\partial y}=\frac{{\partial }^{2}h}{\partial y\partial x}.$

$\frac{{\partial }^{2}h}{\partial x\partial y}=\frac{{\partial }^{2}h}{\partial y\partial x}$

See the step by step solution

## Step 1. Given information

$h\left(x,y\right)=f\left(x\right)g\left(y\right)\phantom{\rule{0ex}{0ex}}\mathrm{Where},f\left(x\right)is\mathrm{a}\mathrm{differentiable}\mathrm{function}\mathrm{of}x,g\left(y\right)is\mathrm{a}\mathrm{differentiable}\mathrm{function}\mathrm{of}y.$

## Step 2. Proof of given partial derivative

$\mathrm{LHS}=\frac{{\partial }^{2}h}{\partial x\partial y}\phantom{\rule{0ex}{0ex}}=\frac{\partial }{\partial x}\left(\frac{\partial f\left(x\right)g\left(y\right)}{\partial y}\right)\phantom{\rule{0ex}{0ex}}=\frac{\partial f\left(x\right)g\text{'}\left(y\right)}{\partial x}\phantom{\rule{0ex}{0ex}}=f\text{'}\left(x\right)g\text{'}\left(y\right)\phantom{\rule{0ex}{0ex}}\mathrm{RHS}=\frac{{\partial }^{2}h}{\partial y\partial x}\phantom{\rule{0ex}{0ex}}=\frac{\partial }{\partial y}\left(\frac{\partial f\left(x\right)g\left(y\right)}{\partial x}\right)\phantom{\rule{0ex}{0ex}}=\frac{\partial f\text{'}\left(x\right)g\left(y\right)}{\partial y}\phantom{\rule{0ex}{0ex}}=f\text{'}\left(x\right)g\text{'}\left(y\right)\phantom{\rule{0ex}{0ex}}\mathrm{LHS}=\mathrm{RHS}$ ### Want to see more solutions like these? 