help with logarithmic differentiation

[yaowangatom]

can someone explain why the professor took 4 points out of 10 from me?

 

[Deleted member 4993]

can someone explain why the professor took 4 points out of 10 from me?

I do not see anything wrong "mathematically" with your work of calculating f'(x).

However, the question is asking you to evaluate f(x) - and not f'(x). That is a bit strange, but that is what is written for instruction.

So the instructor may have had a different requirement for presentation of your work. You would need to ask the instructor.

 

[yaowangatom]

I do not see anything wrong "mathematically" with your work of calculating f'(x).

However, the question is asking you to evaluate f(x) - and not f'(x). That is a bit strange, but that is what is written for instruction.

So the instructor may have had a different requirement for presentation of your work. You would need to ask the instructor.

thanks, professor said he would give us another chance to redo it. if we do it right, he would give us the point back……

 

[yaowangatom]

I do not see anything wrong "mathematically" with your work of calculating f'(x).

However, the question is asking you to evaluate f(x) - and not f'(x). That is a bit strange, but that is what is written for instruction.

So the instructor may have had a different requirement for presentation of your work. You would need to ask the instructor.

evaluating f(x) is not same as looking for f’(x)?

 

[Ishuda]

can someone explain why the professor took 4 points out of 10 from me?

As Subhotosh Khan said, I can't see anything wrong 'mathematically' with your work. The only thing I could think of which might apply is that the answer can be written in 'simpler' form
f'(x) = [ f(x) / x - 2 / (x+1) ] f(x)
but, as far as I'm concerned, that would really be stretching it.

 

[pka]

can someone explain why the professor took 4 points out of 10 from me?

\(\displaystyle 2\left[ {\log \left( {1 + \frac{1}{x}} \right) - \frac{1}{{x + 1}}} \right]{\left( {1 + \frac{1}{x}} \right)^{2x}}\)
You certainly got the correct answer but went around the barn twice to get it. For that I would not have docked you 40%. Maybe 10%??

Look at this:
\(\displaystyle \begin{array}{l}y=\log {\left( {1 + \frac{1}{x}} \right)^{2x}}\\y=2x\log \left( {1 + \frac{1}{x}} \right)\\y'=2\log \left( {1 + \frac{1}{x}} \right) + \frac{{2x\left( { - {x^{ - 2}}} \right)}}{{\left( {1 + \frac{1}{x}} \right)}}\\y'=2\left[ {\log \left( {1 + \frac{1}{x}} \right) - \frac{1}{{x + 1}}} \right]\end{array}\)

 

[stapel]

evaluating f(x) is not same as looking for f’(x)?

No; plugging a specific alue in for "x" and finding the specific corresponding value of y = f(x) is not at all the same thing as differentiating the function to find the general derivative. :shock:

 

[Steven G]

can someone explain why the professor took 4 points out of 10 from me?

The wording of this problem is simply terrible!