por

Como você diferencia # y = ln (1 + x ^ 2) #?

Responda:

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

Explicação:

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)#