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Chapter 11: Problem 106

Consider the function \(f(x, y)=\left(x^{2}+y^{2}\right)^{2 / 3}\). Show that \(f_{x}(x, y)=\left\\{\begin{array}{ll}\frac{4 x}{3\left(x^{2}+y^{2}\right)^{1 / 3}}, & (x, y) \neq(0,0) \\ 0, & (x, y)=(0,0)\end{array}\right.\)

Short Answer

Expert verified
So, \(f_{x}(x, y)=\left\{\begin{array}{ll}\frac{4x}{3(x^{2}+y^{2})^{1/3}}, & (x, y) \neq(0,0) \ 0, & (x, y)=(0,0)\end{array}\right.\)

Step by step solution

01

Understand the Function

The function \(f(x, y)=\left(x^{2}+y^{2}\right)^{2 / 3}\) given is a multivariable function with two variables 'x' and 'y'.
02

Compute the Partial Derivative of Function

To compute \(f_{x}(x, y)\), which is the partial derivative of the function with respect to 'x', differentiate the function partially with 'x'. So, \(f_{x}(x, y)=\frac{d}{dx}\left[(x^{2}+y^{2})^{2 / 3}\right]\). Applying the chain rule of derivatives, we get \(f_{x}(x, y)= \frac{2}{3}(x^{2}+y^{2})^{-1/3} * 2x = \frac{4x}{3(x^{2}+y^{2})^{1/3}}\)
03

Consider the Boundary Condition

The boundary condition given is (x, y) = (0,0). Substituting these values in the partially derived function, we get \(f_{x}(0, 0)= 0\). This is because the numerator becomes zero.

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Most popular questions from this chapter

In Exercises \(83-86,\) show that the function is differentiable by finding values for \(\varepsilon_{1}\) and \(\varepsilon_{2}\) as designated in the definition of differentiability, and verify that both \(\varepsilon_{1}\) and \(\varepsilon_{2} \rightarrow 0\) as \((\boldsymbol{\Delta x}, \boldsymbol{\Delta} \boldsymbol{y}) \rightarrow(\mathbf{0}, \mathbf{0})\) \(f(x, y)=x^{2}-2 x+y\)

Find \(d w / d t\) (a) using the appropriate Chain Rule and (b) by converting \(w\) to a function of \(t\) before differentiating. \(w=x y \cos z, \quad x=t, \quad y=t^{2}, \quad z=\arccos t\)

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