Logarithmic Differentiation equation, Help! So, I have to differentiate this via $\log$. I am still learning, so please be patient, I will try to explain everything I did. Please tell me if it is correct.
$$y=\frac{(x+3)^4(2x^2+5x)^3}{\sqrt{4x-3}}$$
$$\ln(y) = \ln\left((x+3)^4(2x^2+5x)^3)\right) - \ln\left(\sqrt{4x-3}\right)$$
$$\ln(y) = 4\ln(x+3) 3\ln(2x^2+5x) - \frac{1}{2} \ln(4x-3)$$
aaaaaaaaand don't know what to do next, any help in the process or next step?
 A: taking th derivative of the last equation you have, we get $$\frac1y\frac{dy}{dx } = \frac{4}{x+3} + \frac{3(4x+5)}{2x^2 + 5x}-\frac{2}{4x-3} $$
A: You've left out a plus sign in the last line. It should read
$$\ln(y) = 4\ln(x+3) + 3\ln(2x^2+5x) - \frac{1}{2} \ln(4x-3).$$
Then you can still factor the second term to get
\begin{align*}
\ln(y) &= 4\ln(x+3) + 3\ln(x(2x+5)) - \frac{1}{2} \ln(4x-3)\\
&=4\ln(x+3) + 3\ln(x)+3\ln(2x+5) - \frac{1}{2} \ln(4x-3).
\end{align*}
From that point, you can go ahead and differentiate to obtain
$$\frac{y'}{y}=\frac{4}{x+3}+\frac{3}{x}+\frac{6}{2x+5}-\frac{2}{4x-3}.$$
Then multiply the $y$ across to get
$$y'=\left(\frac{4}{x+3}+\frac{3}{x}+\frac{6}{2x+5}-\frac{2}{4x-3}\right)y.$$
Finally, you would like to express $y'$ as a function of $x$, rather than both $x$ and $y$, so substitute in your original equation
$$y=\frac{(x+3)^4(2x^2+5x)^3}{\sqrt{4x-3}}$$
on the right to obtain
$$y'=\left(\frac{4}{x+3}+\frac{3}{x}+\frac{6}{2x+5}-\frac{2}{4x-3}\right)\left(\frac{(x+3)^4(2x^2+5x)^3}{\sqrt{4x-3}}\right),$$
 and then simplify to taste.
