How do you find dy/dx given #y=ln(2+x^2)#?
1 Answer
May 12, 2016
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)#
