How Can You Simplify the Derivative of Ln(x + 1/x)?

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The discussion focuses on simplifying the derivative of the function Ln(x + 1/x). The initial differentiation approach involves applying the chain rule, resulting in the expression (x/(x^2 + 1)) * (1 - 1/x^2). Participants suggest rewriting the expression into a single fraction and utilizing partial fractions to decompose terms. The final goal is to express the derivative in the form (2x/(x^2 + 1)) - (1/x), which requires careful manipulation of the terms.

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jorgen
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Hi all,

I have to differentiate

Ln(x+\frac{1}{x})

where I first differentiate Ln and than multiply by the differentiation of the inner function

1/(x+\frac{1}{x})*(1-\frac{1}{x^2}

which I simplify to

\frac{x}{x^2+1}*(1-\frac{1}{x^2})

\frac{x}{x^2+1}-\frac{1}{x*(x^2+1)}

the problem is I cannot rewrite it to this

\frac{2*x}{x^2+1}-\frac{1}{x}

how to rewrite it - any help or advise appreciated. Thanks in advance

Best
Jorgen
 
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Start at the final expression, and write it as a fraction with one denominator: x(x^2+1)
 
thanks,

so I put into one fraction

\frac{x^2-1}{x*(x^2+1)}

but I don't know how to start rearranging this... Any new hints

Best

J
 
x^2-1=(x^2+1)-2 ;0)
 
so I rewrite the fraction using this hint

\frac{(x^2+1)-2}{x*(x^2+1)}

I split the fraction into

\frac{1}{x}-\frac{2}{x*(x^2+1)}

but I can still not see how to rearrange it

Thanks in advance

Best
J
 
Use partial fractions to decompose your second term; in other words, find the constants A,B, and C that satisfy:

\frac{2}{x*(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}
 
In what form do you want it? Whether it is condensed into one fraction or written as a difference doesn't matter if both expressions are equal.
 

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