Question

How do you find dy/dx given `y=ln(2+x^2)`?

Answer 1

`(2x)/(2+x^2)`

Explanation:

Using `d/dx(ln(f(x)))=1/f(x)`

combined with the `color(blue)" chain rule"`

`d/dx[f(g(x))]=f'(g(x)).g'(x)color(green)" A"`
`"------------------------------------------------"`

here `f(g(x))=ln(2+x^2)rArrf'(g(x))=1/(2+x^2)`

and `g(x)=2+x^2rArrg'(x)=2x`
`"---------------------------------------------------------------"`
Substitute these values into`color(green)" A"`

`rArrdy/dx=1/(2+x^2) xx2x=(2x)/(2+x^2)`