Question

Logarithmic differantiation, y = (x^3 -2x)^ln(x)

I got a very long result I dont know if my solution is correct

Answer 1

y'=(x^3 - 2x)^ln(x)lnx^3-2x+lnx*(2x-2)/(x^3-2x)

Answer 2

I forgot an 1/x next to (x^3 - 2x)

Answer 3

multiplication

Answer 4

I got\[\Large y'=[\frac{(3x^2-2)\ln(x)}{x^3-2x}+\frac{\ln(x^3-2x)}{x}](x^3-2x)^{\ln(x)}\]

Answer 5

hm

Answer 6

only thing to correct in your answer: it's \(3x^2-2\) instead of \(2x-2\), and you forgot some parentheses.

Answer 7

Yea I just noticed that error, and the parenthesis I know, I just surpassed them. Other than that all ok ??

Answer 8

yes

Answer 9

Yeah
\[y=(x^3-2x)^{\ln(x)} \\ \\ \ln(y)=\ln(x^3-2x)^{\ln(x)} \\ \\ \ln(y)=\ln(x)\ln(x^3-2x)\]
Implicit differentiation , and product rule + chain rule for RHS
\[\frac{1}{y}y'=\ln(x)\frac{1}{x^3-2x}(3x^2-2)+\ln(x^3-2x)(\frac{1}{x})\]

Answer 10

thank you guys