Using logarithmic differentiation, find $y'$. Am I correct?

Asked Jun 11 '21 at 12:44

Active Jun 11 '21 at 13:18

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Using logarithmic differentiation, find $y'$.

$$y=\ln x^{\cos(x)} \space (x>1)$$

I solved and here is my answer:

$$y'=(\ln x)^{\cos(x)}(\cfrac{\cos(x)}{x\ln x}-\sin x\ln(\ln x)) $$

But I don't understand what is $(x>1)$ use for?

Or am I missed something?

edited Jun 11 '21 at 13:18

Aaron Hendrickson

asked Jun 11 '21 at 12:44

2

Note that $\ln(x)$ is negative for $x<1$ and $\cos(x)$ is not an integer for the same range, so $\ln(x)^\cos(x)$ makes no sense for $x<1$. – Nah Jun 11 '21 at 12:47
Then, is my answer correct? – Jun 11 '21 at 13:04
Yes, it is correct. – Nah Jun 11 '21 at 13:06
Thank you so much. – Jun 11 '21 at 13:06
You should probably adjust your original equation to read $$(\ln(x))^{\cos(x)}.$$ The way it's written is a bit ambiguous and you fixed it down below anyway. – Cameron Williams Jun 11 '21 at 13:17
@CameronWilliams Okay. – Jun 11 '21 at 13:19

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