# a) Find f_{xy}(x,y); ((4x^{2})/y) + ((y^{2})/(2x)). b) Find f_{xy}(x,y); ln(x^{2} - y^{2}). c)...

## Question:

a) Find {eq}f_{xy}(x,y); \, \frac{4x^{2}}{y} + \frac{y^{2}}{2x}. {/eq}

b) Find {eq}f_{xy}(x,y); \, \ln(x^{2} - y^{2}). {/eq}

c) Find {eq}\frac{dw}{dt} {/eq} using the chain rule.

{eq}w = \ln(xy) + xy^2; \, x = e^{t}; \, y = e^{-t} {/eq}

## Chain Rule:

Suppose {eq}x=x\left( t \right) {/eq} and {eq}y=y\left( t \right) {/eq} and {eq}w=f\left( x,y \right) {/eq}, then

{eq}\frac{dw}{dt}=\frac{dw}{dx}.\frac{dx}{dt}+\frac{dw}{dy}.\frac{dy}{dt} {/eq}

Similarly, if we have any function of three or more variables, we can find the derivative using chain rule.

Partial derivatives of a multi-variable function are calculated the same way as we differentiate a single variable function. We hold all variables except one constant.

## Answer and Explanation: 1

Part (a):

{eq}\begin{align} & f\left( x,y \right)=\frac{4{{x}^{2}}}{y}+\frac{{{y}^{2}}}{2x} \\ & f\left( x,y \right) =\frac{4{{x}^{2}}}{y}+\frac{{{y}^{2}}}{2}{{x}^{-1}} \\ \end{align} {/eq}

Take partial derivative w.r.t. {eq}x {/eq}. Use power rule of differentiation

{eq}\begin{align} & {{f}_{x}}\left( x,y \right)=\frac{8x}{y}-\frac{{{y}^{2}}}{2}{{x}^{-2}} \\ & {{f}_{x}}\left( x,y \right) =\frac{8x}{y}-\frac{{{y}^{2}}}{2{{x}^{2}}} \\ \end{align} {/eq}

Now do partial differentiation w.r.t. {eq}y {/eq}

{eq}\begin{align} & {{f}_{x}}\left( x,y \right)=8x{{y}^{-1}}-\frac{{{y}^{2}}}{2{{x}^{2}}} \\ & {{f}_{xy}}\left( x,y \right)=-8x{{y}^{-2}}-\frac{2y}{2{{x}^{1}}} \\ & {{f}_{xy}}\left( x,y \right) =\boxed{\frac{-8x}{{{y}^{2}}}-\frac{y}{{{x}^{2}}}} \\ \end{align} {/eq}

{eq}\; {/eq}

Part (b):

{eq}f\left( x,y \right)=\ln \left( {{x}^{2}}-{{y}^{2}} \right) {/eq}

Take partial derivative w.r.t. {eq}x {/eq}. Use chain rule {eq}\left( \frac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)g'\left( x \right) \right) {/eq} and {eq}\frac{d}{dx}\left[ \ln \left( x \right) \right]=\frac{1}{x} {/eq}

{eq}\begin{align} & {{f}_{x}}\left( x,y \right)=\frac{\partial }{\partial x}\left( \ln \left( {{x}^{2}}-{{y}^{2}} \right) \right) \\ & {{f}_{x}}\left( x,y \right) =\left( \frac{1}{{{x}^{2}}-{{y}^{2}}} \right).\frac{\partial }{\partial x}\left( {{x}^{2}}-{{y}^{2}} \right) \\ & {{f}_{x}}\left( x,y \right) =\frac{2x}{{{x}^{2}}-{{y}^{2}}} \\ \end{align} {/eq}

Now take partial derivative w.r.t. {eq}y {/eq}. Use power rule and chain rule

{eq}\begin{align} & {{f}_{x}}\left( x,y \right)=2x{{\left( {{x}^{2}}-{{y}^{2}} \right)}^{-1}} \\ & {{f}_{xy}}\left( x,y \right)=-2x{{\left( {{x}^{2}}-{{y}^{2}} \right)}^{-2}}.\frac{\partial }{\partial y}\left( {{x}^{2}}-{{y}^{2}} \right) \\ & {{f}_{xy}}\left( x,y \right)=\frac{\left( -2x \right)\left( -2y \right)}{{{\left( {{x}^{2}}-{{y}^{2}} \right)}^{2}}} \\ & {{f}_{xy}}\left( x,y \right)=\boxed{\frac{4xy}{{{\left( {{x}^{2}}-{{y}^{2}} \right)}^{2}}}} \\ \end{align} {/eq}

{eq}\; {/eq}

Part (c):

{eq}\begin{align} & w=\ln \left( xy \right)+x{{y}^{2}} \\ & x={{e}^{t}} \\ & y={{e}^{-t}} \\ \end{align} {/eq}

Using chain rule for two variable function

{eq}\begin{align} & \frac{dw}{dt}=\frac{dw}{dx}.\frac{dx}{dt}+\frac{dw}{dy}.\frac{dy}{dt} \\ & \frac{dw}{dt}=\frac{d}{dx}\left[ \ln \left( xy \right)+x{{y}^{2}} \right].\frac{d}{dt}\left( {{e}^{t}} \right)+\frac{d}{dy}\left[ \ln \left( xy \right)+x{{y}^{2}} \right].\frac{d}{dt}\left( {{e}^{-t}} \right) \\ \end{align} {/eq}

Again use the rule {eq}\frac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right).g'\left( x \right) {/eq}

{eq}\begin{align} & \frac{dw}{dt}=\left[ \frac{1}{xy}.\frac{d}{dx}\left( xy \right)+{{y}^{2}} \right]{{e}^{t}}+\left[ \frac{1}{xy}\frac{d}{dy}\left( xy \right)+2xy \right]\left( -{{e}^{-t}} \right) \\ &\frac{dw}{dt}=\left( \frac{1}{xy}.y+{{y}^{2}} \right){{e}^{t}}+\left( \frac{1}{xy}x+2xy \right)\left( -{{e}^{-t}} \right) \\ & \frac{dw}{dt}=\boxed{\left( \frac{1}{x}+{{y}^{2}} \right){{e}^{t}}-\left( \frac{1}{y}+2xy \right){{e}^{-t}}} \\ \end{align} {/eq}

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from

Chapter 14 / Lesson 4This lesson defines the chain rule. It goes on to explore the chain rule with partial derivatives and integrals of partial derivatives.