- #1
amb1989
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Is this a u-sub? I went through and got 1/(2x)^2 but I am not sure if that is correct.
danago said:You can use a substitution, but the answer is not quite 1/(2x)^2.
How did you arrive at that answer?
danago said:Hmm I am not really sure how you ended up xln(x)-(integral sign)(1)(1/(2x)^2).
I would choose the same parts as you did, i.e. u = (ln x)^2 and dv = 1, but I am not sure you applied the formula correctly.
[tex]\int u dv =uv - \int v du[/tex]
Is that what you are using?
The derivative of (ln(x))^2 is 2ln(x)/x.
To find the derivative of (ln(x))^2, you can use the power rule and the chain rule. First, rewrite the expression as (ln(x))^2 = (ln(x))*(ln(x)). Then, using the power rule, the derivative is (ln(x))'*(ln(x)) + (ln(x))*(ln(x))' = 2ln(x)/x.
The derivative of (ln(x))^2 is 2ln(x)/x because of the power rule and the chain rule. The power rule states that the derivative of x^n is nx^(n-1). And the chain rule states that the derivative of f(g(x)) is f'(g(x))*g'(x). Applying these rules to (ln(x))^2, we get 2ln(x)/x.
Yes, the derivative of (ln(x))^2 can be simplified further. Using logarithm rules, we can rewrite 2ln(x)/x as 2ln(x^(1/x)) = 2(ln(x)/x)ln(x). This may be a more simplified form, depending on the context of the problem.
Yes, there are many real-life applications of the derivative of (ln(x))^2. One example is in economics and business, where it can be used to calculate marginal cost and marginal revenue. It is also used in statistics and probability to calculate the probability density function of certain distributions. In physics, it is used to calculate rates of change in various systems.