Please help with a calculus question

[mai4213]

http://i289.photobucket.com/albums/ll211/zephyrxX_Y/ScreenShot2012-11-12at82432PM.png

The link leads to the question.

To be honest, I actually don't know how to start doing this question. Please help

Thank you T_T

 

[girodd]

http://i289.photobucket.com/albums/ll211/zephyrxX_Y/ScreenShot2012-11-12at82432PM.png

The link leads to the question.

To be honest, I actually don't know how to start doing this question. Please help

Thank you T_T

Your starting point is logarithmic differentiation, d/dx (ln(f(x)) = f'(x)/f(x)

 

[soroban]

Hello, mai4213!

I don't agree with their derivative.

56. Consider the function: .\(\displaystyle f(x) \:=\:\frac{1}{x^x}\)

(a) Show that: .\(\displaystyle f'(x) \:=\:f(x)\left(\frac{1-\ln x}{x^2}\right)\) .??

We have: .\(\displaystyle f(x) \:=\: x^{-x}\)


Take logs: .\(\displaystyle \ln\big(f(x)\big) \:=\:\ln(x^{-x}) \:=\: -x\!\cdot\!\ln x\)


Differentiate implicitly:
.. \(\displaystyle \dfrac{1}{f(x)}\cdot f'(x) \:=\:-1\!\cdot\!\ln x - x\!\cdot\!\frac{1}{x} \:=\: -\ln x - 1 \:=\: -(1 + \ln x)\)


Therefore: .\(\displaystyle f'(x) \;=\;-f(x)(1 + \ln x)\)

 

[girodd]

Hello, mai4213!

I don't agree with their derivative.


We have: .\(\displaystyle f(x) \:=\: x^{-x}\)


Take logs: .\(\displaystyle \ln\big(f(x)\big) \:=\:\ln(x^{-x}) \:=\: -x\!\cdot\!\ln x\)


Differentiate implicitly:
.. \(\displaystyle \dfrac{1}{f(x)}\cdot f'(x) \:=\:-1\!\cdot\!\ln x - x\!\cdot\!\frac{1}{x} \:=\: -\ln x - 1 \:=\: -(1 + \ln x)\)


Therefore: .\(\displaystyle f'(x) \;=\;-f(x)(1 + \ln x)\)

​I think it is likely that the problem is misstated. Things work out as expected if f(x) = x^(1/x) rather than (1/x)^x.