Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.3 Exercises - Page 896: 50

Answer

$$\frac{\partial f(1,1)}{\partial x}=\frac{10}{27}$$ $$\frac{\partial f(1,1)}{\partial y}=\frac{8}{27}$$

Work Step by Step

We will use quotient rule to find both partial derivatives. The partial derivative with the respect to $x$ is: $$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}\left(\frac{2xy}{\sqrt{4x^2+5y^2}}\right)=\frac{\sqrt{4x^2+5y^2}\frac{\partial }{\partial x}(2xy)-2xy\frac{\partial }{\partial x}(\sqrt{4x^2+5y^2})}{(\sqrt{4x^2+5y^2})^2}=\frac{\sqrt{4x^2+5y^2}\cdot2y-2xy\cdot\frac{1}{2\sqrt{4x^2+5y^2}}\frac{\partial }{\partial x}(4x^2+5y^2)}{4x^2+5y^2}=\frac{(4x^2+5y^2)\cdot2y-xy\cdot8x}{(\sqrt{4x^2+5y^2})^3}=\frac{8x^2y+10y^3-8x^2y}{(\sqrt{4x^2+5y^2})^3}=\frac{10y^3}{(\sqrt{4x^2+5y^2})^3}$$ because $y$ is treated as a constant. We alse used chain rule to find $\frac{\partial }{\partial x}(\sqrt{4x^2+5y^2})$. The partial derivative with the respect to $y$ is: $$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}\left(\frac{2xy}{\sqrt{4x^2+5y^2}}\right)=\frac{\sqrt{4x^2+5y^2}\frac{\partial }{\partial y}(2xy)-2xy\frac{\partial }{\partial y}(\sqrt{4x^2+5y^2})}{(\sqrt{4x^2+5y^2})^2}= \frac{\sqrt{4x^2+5y^2}\cdot2x-2xy\cdot\frac{1}{2\sqrt{4x^2+5y^2}}\frac{\partial }{\partial y}(4x^2+5y^2)}{4x^2+5y^2}=\frac{(4x^2+5y^2)\cdot2x-xy\cdot10y}{(\sqrt{4x^+5y^2})^3}=\frac{8x^3+10y^2x-10y^2x}{(\sqrt{4x^2+5y^2})^3}=\frac{8x^3}{(\sqrt{4x^2+5y^2})^3}$$ because now $x$ is treated as a constant. Again, we used the chain rule to find $\frac{\partial}{\partial y}(\sqrt{4x^2+5y^2}).$ Now we will evaluate these partial derivatives in the given point $(x,y)=(1,1)$: $$\frac{\partial f(1,1)}{\partial x}=\left.\frac{10y^3}{(\sqrt{4x^2+5y^2})^3}\right|_{(x,y)-(1,1)}=\frac{10\cdot1^3}{(\sqrt{4\cdot1^2+5\cdot1^2})^3}=\frac{10}{(\sqrt9)^3}=\frac{10}{27}$$ $$\frac{\partial f(1,1)}{\partial y}=\left.\frac{8x^3}{(\sqrt{4x^2+5y^2})^3}\right|_{(x,y)=(1,1)}=\frac{8\cdot1^3}{(\sqrt{4\cdot1^2+5\cdot1^2})^3}=\frac{8}{(\sqrt9)^3}=\frac{8}{27}$$
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