Answer:
[tex]\frac{dy}{dx}= \frac{2x}{x^2+1}[/tex]
Step-by-step explanation:
Use the property :
[tex]\frac{d}{dx}(ln\ x)=\frac{1}{x}[/tex]
Lets get started
[tex]y=ln\ (2x^2+2)\\\\\frac{dy}{dx} =\frac{1}{2x^2+2} (\frac{d}{dx}(2x^2+2) )\\\\\frac{dy}{dx}= \frac{1}{2x^2+2}(4x+0)\\\\\frac{dy}{dx}= \frac{1}{2x^2+2}(4x)\\\\\frac{dy}{dx}= \frac{1}{x^2+1}(2x)\\\\\frac{dy}{dx}= \frac{2x}{x^2+1}[/tex]
