Question

Evaluate the second partial derivatives $f_{x{x}^{\prime }}{f}_{x{y}^{\prime }}{f}_{y{y}^{\prime }}$ and $f_{yx}$ at the point. $$f(x,y)=\sqrt{x^2+y^2}$$

Short answer

The second order partial derivatives of $f(x,y)=\sqrt{x^2+y^2}$ are $f_{xx}=\frac{y^2}{(x^2+y^2{)}^{3/2}},f_{yy}=\frac{x^2}{(x^2+y^2{)}^{3/2}},f_{xy}=f_{yx}=\frac{-xy}{(x^2+y^2{)}^{3/2}}$.

Step-by-step solution

Compute the first order partial derivatives

First, find the first-order partial derivatives with respect to x and y. They are given by $f_x=\frac{x}{\sqrt{x^2+y^2}}$ and $f_y=\frac{y}{\sqrt{x^2+y^2}}$ respectively.

Compute the second order partial derivatives

Now compute the second order partial derivatives by differentiating $f_x$ and $f_y$ again with respect to x and y respectively. They are given by $f_{xx}=\frac{y^2}{(x^2+y^2{)}^{3/2}}$ , $f_{yy}=\frac{x^2}{(x^2+y^2{)}^{3/2}}$ and since derivative is commutative $f_{xy}=f_{yx}=\frac{-xy}{(x^2+y^2{)}^{3/2}}$.

Evaluate the partial derivatives at the point

The student hasn't provided a specific point, so the second order derivatives are left in terms of x and y. If a specific point is provided, subsitute their coordinates into each of these derived expressions.

Key concepts

Partial Derivatives Calculus

Partial derivatives are a fundamental concept in calculus when dealing with functions of multiple variables. They measure how a function changes as only one of the variables is altered, holding the others constant.

Imagine you are walking on a hilly terrain, the slope (steepness) of the hill under your feet might change if you walk north versus if you walk east. Partial derivatives would give you the rate of this change for each direction separately.

When computing partial derivatives, you treat all variables except the one you're differentiating with respect to as constants. So, for a function like $f(x,y)=\sqrt{x^2+y^2}$, when you find $f_x$, the partial derivative with respect to $x$, you consider $y$ to be a constant, and differentiate $f$ just like you would with a one-variable function. This process leads to the first order partial derivatives, which then can be further differentiated to find even higher order derivatives, such as the second order ones.

Multivariable Calculus

In multivariable calculus, functions depend on more than one input, meaning they have several variables to consider. Our example function, $f(x,y)=\sqrt{x^2+y^2}$, is a two-variable function often encountered in problems involving distance from a point to the origin in a coordinate system, illustrating how functions can represent physical quantities.

The computation of partial derivatives is integral to multivariable calculus. After you've calculated the first order derivatives, common tasks include finding the gradient vector, which includes all the first order partial derivatives and represents the direction of steepest ascent, or computing higher order derivatives like the second order ones to assess the curvature of the function's graph.

Importance of Second Order Partial Derivatives

Second order partial derivatives can tell us more about the function’s behavior. They can indicate concavity and points of inflection, and are essential in optimization problems where you might want to find the maximum or minimum values of a function.

Derivative Computation

Derivative computation is the method by which we determine the rate at which a function changes. In the context of multivariable functions, we perform this computation for each variable independently. The chain rule, product rule, quotient rule, and other familiar techniques from single-variable calculus still apply, but they're used in a more nuanced way due to the presence of multiple variables.

For our sample function, $f(x,y)$, we used these rules to find the first and second order partial derivatives. The calculation of $f_{xx}$, which required differentiating $f_x$ with respect to $x$ again, showcases derivative computation in action. This second order derivative tells us how the slope along the x-direction changes as we move in the x-direction. Similar computations are done for $f_{yy}$ and $f_{xy}$. Because of the symmetry in partial differentiation, $f_{xy}$ and $f_{yx}$ are equal due to Clairaut's theorem, making the computations efficient.