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

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

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

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

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

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

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