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?
On
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.
taking th derivative of the last equation you have, we get $$\frac1y\frac{dy}{dx } = \frac{4}{x+3} + \frac{3(4x+5)}{2x^2 + 5x}-\frac{2}{4x-3} $$