## 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

1. Christos

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

2. Christos

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

3. Christos

multiplication

4. .Sam.

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

5. Christos

hm

6. reemii

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

7. Christos

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

8. reemii

yes

9. .Sam.

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 $\frac{1}{y}y'=\ln(x)\frac{1}{x^3-2x}(3x^2-2)+\ln(x^3-2x)(\frac{1}{x})$

10. Christos

thank you guys