Question

y=(lnx^2)/(ln⁡(x^2-1)), find Y'

Answer 1

\[\frac{\textrm{d}}{\textrm{d}x}\left[\frac{f(x)}{g(x)}\right]
=\frac{g(x)f'(x)-f(x)g'(x)}{{\textbf{[}g(x)\textbf{]}}^{\large2}}\\
y=\frac{\ln x^{\large2}}{\ln{(x^{\large2}-1)}} \\
\begin{align}
y' &= \frac{\ln(x^{\large2}-1)\cdot\dfrac{1}{\ln x^{\large2}}\cdot2x-
\ln x^{\large2}\cdot\dfrac{1}{\ln (x^{\large2}-1)}\cdot2x}{(\ln(x^{\large2}-1))^{\large2}} \\
&=\dfrac{2x}{\ln(x^{\large2}-1)\cdot\ln x^{\large2}} -
\dfrac{2x\ln x^{\large2}}{(\ln(x^{\large2}-1))^{\large3}} \\
&=\dfrac{x}{\ln(x^{\large2}-1)\cdot\ln x} -
\dfrac{4x\ln x}{(\ln(x^{\large2}-1))^{\large3}} \\
&=\dfrac{x\ln(x^{\large2}-1))^{\large2}}{\ln x\cdot(\ln(x^{\large2}-1))^{\large3}} -
\dfrac{4x(\ln x)^{\large2}}{(\ln(x^{\large2}-1))^{\large3}\cdot\ln x} \\
&=\dfrac{x\ln(x^{\large2}-1))^{\large2} - 4x(\ln x)^{\large2}}
{\ln x\cdot(\ln(x^{\large2}-1))^{\large3}}
\end{align}
\]

I don't think that I got that right after all of that mess, but you get the basic idea, with the Quotient Rule.

Answer 2

yeah thanks alot

Answer 3

the error occurs in the 1st line of simplifying the derivative it should be

\[\frac{\frac{2xln(x^2 -1)}{x^2} - \frac{2xln(x)}{x^2 -1}}{(\ln(x^2 -1))^2}\]

Answer 4

I kinda combined 2 steps into 1. What I simplified to is what you would get if you distributed your denominator of \((g(x))^{\large2}\).

Answer 5

but you log derivatives are incorrect

\[y=\log(f(x))... then ...y' = \frac{f'(x)}{f(x)}\]

Answer 6

Oh shiznuggets, you're right. :b