Answer: The required derivative is [tex]\dfrac{8x^2+18x+9}{x(2x+3)^2}[/tex]
Step-by-step explanation:
Since we have given that
[tex]y=\ln[x(2x+3)^2][/tex]
Differentiating log function w.r.t. x, we get that
[tex]\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [x'(2x+3)^2+(2x+3)^2'x]\\\\\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [(2x+3)^2+2x(2x+3)]\\\\\dfrac{dy}{dx}=\dfrac{4x^2+9+12x+4x^2+6x}{x(2x+3)^2}\\\\\dfrac{dy}{dx}=\dfrac{8x^2+18x+9}{x(2x+3)^2}[/tex]
Hence, the required derivative is [tex]\dfrac{8x^2+18x+9}{x(2x+3)^2}[/tex]
