Question

Find F'(x) of the following function:

Answer 1

\[f:f(x)=\frac{ x ^{2}+3 }{ x }- \ln \left| \frac{ x-2 }{ x } \right|\]

Answer 2

\[\frac{df(x)}{dx}=\frac{(\frac{d}{dx}x^2)+3}{x}+\frac{x^2+3}{(\frac{d}{dx}x)}+\frac{d}{dx}(\ln |\frac{x-2}{x}| )\]
\[\frac{df(x)}{dx}={\frac {2\,x+3}{x}}-{\frac {{x}^{2}+3\,x}{{x}^{2}}}- \left( {x}^{-1}-{ \frac {x-2}{{x}^{2}}} \right) x \left( x-2 \right) ^{-1} \]
\[\frac{df(x)}{dx}={\frac {{x}^{2}-2\,x-2}{x \left( x-2 \right) }} \]

Answer 3

If you need further explanation of how the last terms derivative is computed, just ask. Kinda getting tired of writing in LaTeX...

Answer 4

I do have some troubles with the derivative of an abs value.

Answer 5

Do you have any experience with the signum function? What level calculus is this from? Calculus I?

Answer 6

Yes, it's calculus I and i have experience with the sign study.

Answer 7

\[{\frac {d \left| x \right| }{{\it dx}}}={\frac {x}{ \left| x \right| } }\]
I am having some issues with the piecewise format in LaTeX showing up on here. However, if you take the derivative of the absolute value of x it spits out x divided by the absolute value of x. This is an identity for the signum function.

Either way, the take away is this: In the context of most problems, the absolute value function can just be treated as another set of parenthesis... Unless a sub-interval is specified or you are evaluating the derivative at a particular point. If one is, or you are evaluating at a specific point, if the x value of the interval or value that one is evaluating at is greater than 1, it is positive. If it is less than 1 the output becomes negative.

Answer 8

Well, I should state if you are taking the deriviative of an absolute value function without a specified interval or particular point, the output is always positive. \[\frac{d}{dx}abs(x^2)=2*abs(x)=2x\]
\[\frac{d}{dx}abs(-x^2)=2*abs(-x)=2x\]

Answer 9

So you say that:
\[f:f(x)= \left| x \right|\]
then:
\[f'(x)=\frac{ x }{ \left| x \right| }.x'\]
wich I could say:
if:
\[f:f(x)=\left| x \right| \]
\[f'(x)= sg.x\]

Is what I understood from your explaination.

Answer 10

Yes. Sorry, it took me a while to respond. Trying to work on my own homework, lol.