True because yes yes yes yes yes
Answer: The required derivative is 
Step-by-step explanation:
Since we have given that
![y=\ln[x(2x+3)^2]](https://tex.z-dn.net/?f=y%3D%5Cln%5Bx%282x%2B3%29%5E2%5D)
Differentiating log function w.r.t. x, we get that
![\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}](https://tex.z-dn.net/?f=%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7B1%7D%7B%5Bx%282x%2B3%29%5E2%5D%7D%5Ctimes%20%5Bx%27%282x%2B3%29%5E2%2B%282x%2B3%29%5E2%27x%5D%5C%5C%5C%5C%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7B1%7D%7B%5Bx%282x%2B3%29%5E2%5D%7D%5Ctimes%20%5B%282x%2B3%29%5E2%2B2x%282x%2B3%29%5D%5C%5C%5C%5C%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7B4x%5E2%2B9%2B12x%2B4x%5E2%2B6x%7D%7Bx%282x%2B3%29%5E2%7D%5C%5C%5C%5C%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7B8x%5E2%2B18x%2B9%7D%7Bx%282x%2B3%29%5E2%7D)
Hence, the required derivative is 
the answer is 16.5cm
just add 15 +1.5 and you get 16.5
Substitution
-2x+8=x^2-9x+18
x^2-7x+10=0
Factoring, we get
(x-5)(x-2)=0
x=5, x=2
If x equals 5,
y=-2(5)+8=
-10+8=
-2
Therefore, we derive our first solution:
<h2><u><em>
x=5,</em></u></h2><h2><u><em>
y=-2</em></u></h2>
Now we solve for our second x
If x=2,
y=-4+8=
4
Therefore, we derive our second solution:
<h2><u><em>
x=2</em></u></h2><h2><u><em>
y=4</em></u></h2>
<u><em></em></u>
-Hunter