Question

In(x^2+y^2) = e find derivative by using implicit differentiation

Answer 1

Given the equation:

$ln(x^2+y^2)=e$

You can find its derivative by using Implicit Differentiation. In order to do this, you need to treat "y" as a function of "x".

The steps are:

1. Set up:

$\lbrack ln(x^2+y^2)\rbrack^{\prime}=(e)^{\prime}$

2. Apply the following Derivatives Rules:

$\frac{d}{dx}(lnu(x))=\frac{1}{u(x)}\cdot u^{\prime}(x)$$\frac{d}{dx}(k)=0$

Where "k" is a constant.

$\frac{d}{dx}(x^n)=nx^{n-1}$$\frac{d}{dx}(ux^n)=nu(x)^{n-1}\cdot u^{\prime}(x)$

Then, you get:

$(\frac{1}{x^2+y^2})(x^2+y^2)^{\prime}=0$$(\frac{1}{x^2+y^2})(2x+2y\cdot y^{\prime})=0$$\frac{2x+2yy^{\prime}}{x^2+y^2}=0$

3. Solve for:

$y^{\prime}$

Then:

$2(x+yy^{\prime})=(x^2+y^2)(0)$$y^{\prime}=\frac{-2x}{2y}$$y^{\prime}=-\frac{x}{y}$

Hence, the answer is:

$y^{\prime}=-\frac{x}{y}$