MHB How to Simplify the Derivative of $\frac{1+\ln(t)}{1-\ln(t)}$?

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The differentiation of the function $\frac{1+\ln(t)}{1-\ln(t)}$ was performed correctly, resulting in $\frac{1/t(1-\ln(t))-(-1/t)(1+\ln(t))}{(1-\ln(t))^2}$. However, it was noted that this expression can be simplified further for elegance and clarity. Simplifying derivatives is important not only for aesthetic reasons but also for practical applications, such as finding critical numbers and optimizing functions. Engaging in simplification enhances algebra skills and aids in understanding the function's behavior. Overall, confirming the correctness of the differentiation while emphasizing the importance of simplification is crucial.
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Differentiate the function: $\frac{1+\ln(t)}{1-\ln(t)}$
Is this correct?
$\frac{1/t(1-\ln(t))-(-1/t)(1+\ln(t))}{(1-\ln(t))^2}$
I got none facit in My textbook so I would like to someone confirmed I am doing right.
 
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correct
 
It's correct as far as it goes, but you could certainly simplify it a bit.
 
Opalg said:
It's correct as far as it goes, but you could certainly simplify it a bit.

Yes, this is a good habit to get into, as not only does it improve your algebra skills and just looks more elegant to express results in a simpler more compact form, it is also crucial for finding critical numbers when you are looking at the behavior of the original function or for optimization (finding extrema).
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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