Answer:
[tex]\dfrac{dy}{dx} = \frac{2x-3}{\ln 16(x^2 - 3x)}[/tex]
Step-by-step explanation:
We are given the following in the question:
[tex]y = \ln {16}(x^2 - 3x)[/tex]
We have to find the derivative of the given expression.
[tex]y = \ln 16(x^2 - 3x)\\\text{Using the log propert}\\\\\log_a b = \dfrac{\log b}{\log a}\\\\dfrac{d(x^n)}{dx} = nx^{n-1}\\\\\dfrac{d(\log x)}{dx} = \dfrac{1}{x}\\\\\text{\bold{Differentiating we get}}\\\\\displaystyle\frac{dy}{dx} = \frac{d(\ln 16(x^2-3x))}{dx}\\\\= \frac{1}{16(x^2-3x)})\frac{d(x^2-3x)}{dx}\\\\=\frac{1}{\log 16(x^2-3x)}(2x - 3)\\\\= \frac{2x-3}{\ln 16(x^2 - 3x)}[/tex]
[tex]\dfrac{dy}{dx} = \frac{2x-3}{\ln 16(x^2 - 3x)}[/tex]
