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 
8 > 7 +
Subtract 7 from both sides
1 >
Multiply both sides by 6
6 > x Flip it around so it's easier to read
x < 6
You can graph your answer by drawing an open circle at the 6 and coloring the line to the left. The circle should be open, because x is <em>less than 6</em>, not less than or equal to. You would color to the left to show that x can be anything less than 6.
<h3><u>Solution: </u></h3>
Radius of Cylindrical Pillar , r = 28 Cm = 0.28 m.
Curved surface area of a Cylindrical = 2πrh.
Curved surface area of a pillar -

Curved surface area of 24 such Pillar :-
(7.04 x 24 = 168.96m²)..
- Cost of painting an area of 1 m² = Rs8...
• Therefore , cost of painting 1689.6m² :-
( 168.96 x 8 = Rs 1351.68)...


When

, you're left with

When

or

, you're left with

Adding the two equations together gives

, or

. Subtracting them gives

,

.
Now, you have



By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that

and

. These alone tell you that you must have

and

.
So the partial fraction decomposition is