# Math Help - Differentiate with respect to x

1. ## Differentiate with respect to x

Please could someone show me how to
Differentiate with respect to x

$Y = Ln (x^2+1)/(x^2-1)$

Thanks in advance,

HNC

2. Originally Posted by HNCMATHS
Please could someone show me how to
Differentiate with respect to x

$Y = Ln (x^2+1)/(x^2-1)$

Thanks in advance,

HNC

Quotient Rule

3. Originally Posted by HNCMATHS
Please could someone show me how to
Differentiate with respect to x

$Y = Ln (x^2+1)/(x^2-1)$

Thanks in advance,

HNC

$y=ln\left(\frac{x^2+1}{x^2-1}\right)$

or

$y=\frac{ln(x^2+1)}{x^2-1}\;\;?$

4. $y=ln\left\frac{x^2+1}{x^2-1}\right$

Thanks, sorry for the confusion

5. Originally Posted by HNCMATHS
$y=ln\left\frac{x^2+1}{x^2-1}\right$

Thanks, sorry for the confusion
$\displaystyle\ln\left(\frac{x^2+1}{x^2-1}\right)=\ln(x^2+1)-\ln(x^2-1)$

Use the log rules for differentiation.

6. Originally Posted by dwsmith
$\displaystyle\ln\left(\frac{x^2+1}{x^2-1}\right)=\ln(x^2+1)-\ln(x^2-1)$

Use the log rules for differentiation.
Do you know where I could find them?

7. $\displaystyle\frac{d}{dx}\left[\ln{x}\right]=\frac{\frac{d}{dx}\left[x\right]}{x}$

8. 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

$\ln\bigg(\dfrac{A}{B}\bigg)\;=\;\ln(A)-\ln(B)$

9. Originally Posted by HNCMATHS
$y=ln\left\frac{x^2+1}{x^2-1}\right$

Thanks, sorry for the confusion
$\displaystyle\ u=\frac{x^2+1}{x^2-1}$

$y=ln(u)$

Using the chain rule

$\displaystyle\frac{dy}{dx}=\frac{dy}{du}\;\frac{du }{dx}=\frac{1}{u}\frac{du}{dx}$

Using the Quotient Rule

$\displaystyle\frac{du}{dx}=\frac{\left(x^2-1\right)2x-\left(x^2+1\right)2x}{\left(x^2-1\right)^2}$

$=\displaystyle\frac{2x\left(x^2-x^2-1-1\right)}{\left(x^2-1\right)^2}=-\frac{4x}{\left(x^2-1\right)^2}$

$\displaystyle\frac{dy}{dx}=-\frac{x^2-1}{x^2+1}\;\frac{4x}{\left(x^2-1\right)^2}=-\frac{4x}{\left(x^2+1\right)\left(x^2-1\right)}$

$=\displaystyle\frac{4x}{1-x^4}$

Using log laws

$log\frac{a}{b}=loga-logb$

$log(ab)=loga+logb$

$\displaystyle\ ln\left(\frac{x^2+1}{x^2-1}\right)=ln\left(x^2+1\right)-ln\left(x^2-1\right)$

Using the chain rule

$\displaystyle\frac{dy}{dx}=\frac{1}{x^2+1}(2x)-\frac{1}{x^2-1}(2x)=\frac{\left(x^2-1\right)2x-\left(x^2+1\right)2x}{\left(x^2+1\right)\left(x^2-1\right)}$

$=\displaystyle\frac{2x^3-2x-2x^3-2x}{x^4-1}$

10. Originally Posted by dwsmith
$\displaystyle\ln\left(\frac{x^2+1}{x^2-1}\right)=\ln(x^2+1)-\ln(x^2-1)$

Use the log rules for differentiation.
Adding to this: The derivative of each log term on the right hand side is easily found using the chain rule.

11. $\displaystyle \left[\ln{f(x)}\right]' = \frac{f'(x)}{f(x)}.$

12. Sorry guys, really dont understand this one... I want to use the laws of logs to solve it but i'm not sure how to still

13. Originally Posted by HNCMATHS
Sorry guys, really dont understand this one... I want to use the laws of logs to solve it but i'm not sure how to still
Have you read any of the previous posts?

14. Originally Posted by HNCMATHS
Sorry guys, really dont understand this one... I want to use the laws of logs to solve it but i'm not sure how to still
Use the laws of logs to avoid having to mess with fractions.

You will still need to apply the Chain Rule

and use

$\frac{d}{dx}ln(x)=\frac{1}{x}$

Then

$u=x^2+1\Rightarrow\frac{d}{dx}ln(u)=\frac{d}{du}ln (u)\;\frac{du}{dx}$

so

$\frac{d}{du}ln(u)=\frac{1}{u}$

and so on......

Here's how to use the Chain Rule:

You've gotten to

$ln\left(x^2+1\right)-ln\left(x^2-1\right)$

The derivative "with respect to x" of $ln(x)$ is $\frac{1}{x}$

$\frac{d}{dx}ln(x)=\frac{1}{x}$

We want to differentiate $ln\left(x^2+1\right)$ "with respect to x" however.

If we write $u=x^2+1$ then we can say we are calculating the derivative of $ln(u)$ "with respect to x".

However

$\frac{d}{du}ln(u)=\frac{1}{u}$

$\frac{d}{dx}ln(x)=\frac{1}{x}$

$\frac{d}{dy}ln(y)=\frac{1}{y}$

But what about $\frac{d}{dx}ln(u)\;\;?$

We have to use the Chain Rule for that...

$\frac{d}{dx}ln(u)=\frac{d}{du}ln(u)\;\frac{du}{dx}$

We multiply those 2 derivatives and we can calculate both of them.

$\frac{du}{dx}=\frac{d}{dx}\left(x^2+1\right)=2x$

$\frac{d}{du}ln(u)=\frac{1}{u}$

Now we replace "u" with it's original expression in x.

$\frac{1}{u}=\frac{1}{x^2+1}$

Therefore

$\frac{d}{dx}ln\left(x^2+1\right)=\frac{2x}{x^2+1}$ by multiplying the derivatives.

The same procedure applies to the 2nd log expression. Then the 2 results need to be subtracted.

15. Hi Everyone,

I have done as Archie said, and ended up with...

$=\frac{2x}{x^2+1}$ - $\frac{2x}{x^2-1}$

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,

Page 1 of 2 12 Last