Chapter 3.7, Problem 43E

To determine

To calculate: The derivative of the function $y=\ln \left(x^2+y^2\right)$ .

Answer to Problem 43E

The derivative of the function is $\frac{2x}{x^2+y^2-2y}$ .

Explanation of Solution

Given information:

The function $y=\ln \left(x^2+y^2\right)$ .

Formula used:

Thechain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Product rule for differentiation is $\frac{d}{dx}\left(f\cdot g\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)$ where f and g are functions of x .

Logarithmic rule for differentiation is $\frac{d}{dx}\log x=\frac{1}{x}$ .

Calculation:

Consider the function $y=\ln \left(x^2+y^2\right)$ .

Differentiate both sides with respect to x ,

  $\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\ln \left(x^2+y^2\right)\right)$

Recall that logarithmic rule for differentiation is $\frac{d}{dx}\log x=\frac{1}{x}$ and chain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

  $\begin{array}{c}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\ln \left(x^2+y^2\right)\right)\\ y\text{'}=\frac{1}{x^2+y^2}\cdot \left(2x+2yy\text{'}\right)\\ y\text{'}=\frac{2x+2yy\text{'}}{x^2+y^2}\\ y\text{'}\left(x^2+y^2\right)=2x+2yy\text{'}\end{array}$

Simplify it and isolate y’ on left hand side,

  $\begin{array}{c}y\text{'}\left(x^2+y^2\right)=2x+2yy\text{'}\\ y\text{'}\left(x^2+y^2\right)-2yy\text{'}=2x\\ y\text{'}\left(x^2+y^2-2y\right)=2x\\ y\text{'}=\frac{2x}{x^2+y^2-2y}\end{array}$

Thus, the derivative of the function is $\frac{2x}{x^2+y^2-2y}$ .