Question

How do you differentiate `y=ln(1+x^2)`?

Answer 1

`dy/dx=(2x)/(1+x^2)`

Explanation:

differentiate using the `color(blue)"chain rule"`

That is `color(red)(|bar(ul(color(white)(a/a)color(black)(dy/dx=(dy)/(du)xx(du)/(dx))color(white)(a/a)|)))........ (A)`

`color(orange)"Reminder " color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx(lnx)=1/x)color(white)(a/a)|)))`

let `u=1+x^2rArr(du)/(dx)=2x`

and so `y=lnurArr(dy)/(du)=1/u`

substitute these values into (A) changing u back to terms of x.

`rArrdy/dx=1/u(2x)=(2x)/(1+x^2)`