Answer to Question #321888 in Calculus for Segun

Question #321888

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

Expert's answer

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}

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