Math Problem Statement

Find the partial derivatives of the function f(x, y) = ln(x + sqrt(x^2 + y^2)) with respect to x and y, and evaluate f_x(2,3).

Solution

The given function is:

f(x,y)=ln(x+x2+y2)f(x, y) = \ln\left(x + \sqrt{x^2 + y^2}\right)

1. Partial derivative with respect to xx:

We need to compute the partial derivative fx\frac{\partial f}{\partial x}.

Let’s differentiate step by step:

f(x,y)=ln(x+x2+y2)f(x, y) = \ln\left(x + \sqrt{x^2 + y^2}\right)

Using the chain rule for derivatives: xln(u)=1uux\frac{\partial}{\partial x} \ln(u) = \frac{1}{u} \cdot \frac{\partial u}{\partial x} where u=x+x2+y2u = x + \sqrt{x^2 + y^2}.

Now, we compute ux\frac{\partial u}{\partial x}:

x(x+x2+y2)=1+12(x2+y2)122x\frac{\partial}{\partial x} \left(x + \sqrt{x^2 + y^2}\right) = 1 + \frac{1}{2} \cdot (x^2 + y^2)^{-\frac{1}{2}} \cdot 2x

Simplifying:

x(x+x2+y2)=1+xx2+y2\frac{\partial}{\partial x} \left(x + \sqrt{x^2 + y^2}\right) = 1 + \frac{x}{\sqrt{x^2 + y^2}}

Thus, the partial derivative of ff with respect to xx is:

fx=1x+x2+y2(1+xx2+y2)\frac{\partial f}{\partial x} = \frac{1}{x + \sqrt{x^2 + y^2}} \cdot \left(1 + \frac{x}{\sqrt{x^2 + y^2}}\right)

2. Partial derivative with respect to yy:

Now, we compute the partial derivative fy\frac{\partial f}{\partial y}.

Again, we use the chain rule. First, we differentiate u=x+x2+y2u = x + \sqrt{x^2 + y^2} with respect to yy:

y(x+x2+y2)=12(x2+y2)122y\frac{\partial}{\partial y} \left(x + \sqrt{x^2 + y^2}\right) = \frac{1}{2} \cdot (x^2 + y^2)^{-\frac{1}{2}} \cdot 2y

Simplifying:

y(x+x2+y2)=yx2+y2\frac{\partial}{\partial y} \left(x + \sqrt{x^2 + y^2}\right) = \frac{y}{\sqrt{x^2 + y^2}}

Thus, the partial derivative of ff with respect to yy is:

fy=1x+x2+y2yx2+y2\frac{\partial f}{\partial y} = \frac{1}{x + \sqrt{x^2 + y^2}} \cdot \frac{y}{\sqrt{x^2 + y^2}}

3. Evaluate fx\frac{\partial f}{\partial x} at (2,3)(2, 3):

Substitute x=2x = 2 and y=3y = 3 into the expression for fx\frac{\partial f}{\partial x}:

fx(2,3)=12+22+32(1+222+32)\frac{\partial f}{\partial x} (2, 3) = \frac{1}{2 + \sqrt{2^2 + 3^2}} \cdot \left(1 + \frac{2}{\sqrt{2^2 + 3^2}}\right)

First, compute 22+32=4+9=13\sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}.

Now, substitute:

fx(2,3)=12+13(1+213)\frac{\partial f}{\partial x} (2, 3) = \frac{1}{2 + \sqrt{13}} \cdot \left(1 + \frac{2}{\sqrt{13}}\right)

This is the final value for fx\frac{\partial f}{\partial x} at (2,3)(2, 3).

Would you like a further breakdown or any clarifications?

Related questions:

  1. What is the geometric interpretation of partial derivatives?
  2. How do we apply the chain rule in multivariable functions?
  3. Can you explain how to differentiate logarithmic functions involving multiple variables?
  4. What are some common real-world applications of partial derivatives?
  5. How would you find the second-order partial derivatives of the function?

Tip:

Always simplify the expressions inside the square root or logarithm before differentiating for easier computations.

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Math Problem Analysis

Mathematical Concepts

Partial Derivatives
Logarithmic Functions
Multivariable Calculus

Formulas

f(x, y) = ln(x + sqrt(x^2 + y^2))
∂f/∂x = (1 / (x + sqrt(x^2 + y^2))) * (1 + x / sqrt(x^2 + y^2))
∂f/∂y = (1 / (x + sqrt(x^2 + y^2))) * (y / sqrt(x^2 + y^2))

Theorems

Chain Rule for Partial Derivatives

Suitable Grade Level

Undergraduate Calculus