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

1 Answer
Jim G.
May 12, 2016

#(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)#

Related questions
See all questions in Chain Rule
Impact of this question
2709 views around the world
You can reuse this answer
Creative Commons License