Please could someone show me how to
Differentiate with respect to x
$\displaystyle Y = Ln (x^2+1)/(x^2-1)$
Thanks in advance,
HNC
If you are talking about the identity given by dwsmith, you could find it in your precalc book or look online for properties of logarithms
$\displaystyle \ln\bigg(\dfrac{A}{B}\bigg)\;=\;\ln(A)-\ln(B)$
$\displaystyle \displaystyle\ u=\frac{x^2+1}{x^2-1}$
$\displaystyle y=ln(u)$
Using the chain rule
$\displaystyle \displaystyle\frac{dy}{dx}=\frac{dy}{du}\;\frac{du }{dx}=\frac{1}{u}\frac{du}{dx}$
Using the Quotient Rule
$\displaystyle \displaystyle\frac{du}{dx}=\frac{\left(x^2-1\right)2x-\left(x^2+1\right)2x}{\left(x^2-1\right)^2}$
$\displaystyle =\displaystyle\frac{2x\left(x^2-x^2-1-1\right)}{\left(x^2-1\right)^2}=-\frac{4x}{\left(x^2-1\right)^2}$
$\displaystyle \displaystyle\frac{dy}{dx}=-\frac{x^2-1}{x^2+1}\;\frac{4x}{\left(x^2-1\right)^2}=-\frac{4x}{\left(x^2+1\right)\left(x^2-1\right)}$
$\displaystyle =\displaystyle\frac{4x}{1-x^4}$
Using log laws
$\displaystyle log\frac{a}{b}=loga-logb$
$\displaystyle log(ab)=loga+logb$
$\displaystyle \displaystyle\ ln\left(\frac{x^2+1}{x^2-1}\right)=ln\left(x^2+1\right)-ln\left(x^2-1\right)$
Using the chain rule
$\displaystyle \displaystyle\frac{dy}{dx}=\frac{1}{x^2+1}(2x)-\frac{1}{x^2-1}(2x)=\frac{\left(x^2-1\right)2x-\left(x^2+1\right)2x}{\left(x^2+1\right)\left(x^2-1\right)}$
$\displaystyle =\displaystyle\frac{2x^3-2x-2x^3-2x}{x^4-1}$
Use the laws of logs to avoid having to mess with fractions.
You will still need to apply the Chain Rule
and use
$\displaystyle \frac{d}{dx}ln(x)=\frac{1}{x}$
Then
$\displaystyle u=x^2+1\Rightarrow\frac{d}{dx}ln(u)=\frac{d}{du}ln (u)\;\frac{du}{dx}$
so
$\displaystyle \frac{d}{du}ln(u)=\frac{1}{u}$
and so on......
Here's how to use the Chain Rule:
You've gotten to
$\displaystyle ln\left(x^2+1\right)-ln\left(x^2-1\right)$
The derivative "with respect to x" of $\displaystyle ln(x)$ is $\displaystyle \frac{1}{x}$
$\displaystyle \frac{d}{dx}ln(x)=\frac{1}{x}$
We want to differentiate $\displaystyle ln\left(x^2+1\right)$ "with respect to x" however.
If we write $\displaystyle u=x^2+1$ then we can say we are calculating the derivative of $\displaystyle ln(u)$ "with respect to x".
However
$\displaystyle \frac{d}{du}ln(u)=\frac{1}{u}$
$\displaystyle \frac{d}{dx}ln(x)=\frac{1}{x}$
$\displaystyle \frac{d}{dy}ln(y)=\frac{1}{y}$
But what about $\displaystyle \frac{d}{dx}ln(u)\;\;?$
We have to use the Chain Rule for that...
$\displaystyle \frac{d}{dx}ln(u)=\frac{d}{du}ln(u)\;\frac{du}{dx}$
We multiply those 2 derivatives and we can calculate both of them.
$\displaystyle \frac{du}{dx}=\frac{d}{dx}\left(x^2+1\right)=2x$
$\displaystyle \frac{d}{du}ln(u)=\frac{1}{u}$
Now we replace "u" with it's original expression in x.
$\displaystyle \frac{1}{u}=\frac{1}{x^2+1}$
Therefore
$\displaystyle \frac{d}{dx}ln\left(x^2+1\right)=\frac{2x}{x^2+1}$ by multiplying the derivatives.
The same procedure applies to the 2nd log expression. Then the 2 results need to be subtracted.
Hi Everyone,
I have done as Archie said, and ended up with...
$\displaystyle =\frac{2x}{x^2+1}$ - $\displaystyle \frac{2x}{x^2-1}$
Now how can I simplify that further? I'm not sure how to subtract one from the other as I quite obviouslly need to,
Thanks Again,