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Logarithmic differantiation, y = (x^3 2x)^ln(x)
I got a very long result I dont know if my solution is correct
 10 months ago
 10 months ago
Logarithmic differantiation, y = (x^3 2x)^ln(x) I got a very long result I dont know if my solution is correct
 10 months ago
 10 months ago

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ChristosBest ResponseYou've already chosen the best response.0
y'=(x^3  2x)^ln(x)lnx^32x+lnx*(2x2)/(x^32x)
 10 months ago

ChristosBest ResponseYou've already chosen the best response.0
I forgot an 1/x next to (x^3  2x)
 10 months ago

.Sam.Best ResponseYou've already chosen the best response.0
I got\[\Large y'=[\frac{(3x^22)\ln(x)}{x^32x}+\frac{\ln(x^32x)}{x}](x^32x)^{\ln(x)}\]
 10 months ago

reemiiBest ResponseYou've already chosen the best response.1
only thing to correct in your answer: it's \(3x^22\) instead of \(2x2\), and you forgot some parentheses.
 10 months ago

ChristosBest ResponseYou've already chosen the best response.0
Yea I just noticed that error, and the parenthesis I know, I just surpassed them. Other than that all ok ??
 10 months ago

.Sam.Best ResponseYou've already chosen the best response.0
Yeah \[y=(x^32x)^{\ln(x)} \\ \\ \ln(y)=\ln(x^32x)^{\ln(x)} \\ \\ \ln(y)=\ln(x)\ln(x^32x)\] Implicit differentiation , and product rule + chain rule for RHS \[\frac{1}{y}y'=\ln(x)\frac{1}{x^32x}(3x^22)+\ln(x^32x)(\frac{1}{x})\]
 10 months ago
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