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 
18+21=39
33+49=82
Just add them together
<h2>727</h2>
Step-by-step explanation:
Let us denote the three digit number by
, where
is the hundreds digit,
is the tens digit and
is the ones digit.
Hundreds digit =
= Ones digit = 
Sum of digits = 
So, 
On exchanging tens and ones digit, value increases by 45. So
is 45 greater than
.



∴ Required number is 727
You flip the second fraction and you multiply. 1 and 2/5 or 1.4