Suppose f is a differentiable function of x and y, and g(u, v)=f(4e^u-4sin(v), 4e^u-2cos(v)). Use...


Suppose {eq}f {/eq} is a differentiable function of x and y, and {eq}g(u, v)=f(4e^u-4sin(v), \ 4e^u-2cos(v)) {/eq}. Use the following table of values to calculate {eq}g_u(0, 0) {/eq} and {eq}g_v(0, 0) {/eq}.

f g fx fy
(0, 0) -1 -4 9 12
(4, 2) -15 2 -1 9

Chain Rule for Partial Derivative:

Let us assume two independent variables x and y such that

{eq}\begin{align*} x&\equiv x\left(u,v\right)\\ y&\equiv y\left(u,v\right) \end{align*} {/eq}

It interprets that each of two independent variables x and y is again a function of two newly introduced independent variables u and v.

Another function {eq}g\left(u,v\right) {/eq} is defined by

{eq}g\left(u,v\right)=f\left(x\left(u,v\right),y\left(u,v\right)\right) {/eq}

Then, if we differentiate {eq}g\left(u,v\right) {/eq}w.r.t u and v partially, according to the chain rule of partial derivatives,

{eq}\begin{align*} \frac{\partial g}{\partial u}&=\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}\\ \frac{\partial g}{\partial{v}}&=\frac{\partial f}{\partial x}\frac{\partial x}{\partial{v}}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial{v}}\ \end{align*} {/eq}

At the point {eq}\left(a,b\right) {/eq} , the above partial derivatives become

{eq}\begin{align*} g_u\left(a,b\right)&=f_x\left(a,b\right)x_u\left(a,b\right)+f_y\left(a,b\right)y_u\left(a,b\right)\\ g_v\left(a,b\right)&=f_x\left(a,b\right)x_v\left(a,b\right)+f_y\left(a,b\right)y_v\left(a,b\right) \end{align*} {/eq}.

Answer and Explanation: 1

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Given that

{eq}g\left(u,v\right)=f\left(4e^u-4\sin{v},4e^u-2\cos{v}\right) {/eq}

To calculate {eq}g_u\left(0,0\right) {/eq} and...

See full answer below.

Learn more about this topic:

The Chain Rule for Partial Derivatives


Chapter 14 / Lesson 4

This lesson defines the chain rule. It goes on to explore the chain rule with partial derivatives and integrals of partial derivatives.

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