Answer to Question #321888 in Calculus for Segun

Question

Y= tan^-1(x+1/2x+3)

Answer 1

$$y'=(\tan^{-1}(\dfrac{x+1}{2x+3}))'$$

$$=\dfrac{1}{1+(\dfrac{x+1}{2x+3})^2}(\dfrac{x+1}{2x+3})'$$

$$=\dfrac{(2x+3)^2}{4x^2+12x+9+x^2+2x+1}(\dfrac{2x+3-2(x+1)}{(2x+3)^2})$$

$$=\dfrac{1}{5x^2+14x+10}$$