Taking the derivative of a polynomial fraction?

[mooha]

Taking the derivative of a polynomial fraction??

b]1. Homework Statement [/b]
Ok, so the question wants me to differentiate f(x)= (x)/(x+1). We are supposed to use the definition of the derivative f'(x)= (limit as h->0) [f(x+h)-f(x)]/(h). We also have learned the power rule. I did the formula and had some issues, so I tried to check my answer with the power rule. However they didn't come out the same. I'm not sure where I made the mistake, the formula or the power rule!

2. Homework Equations [/b
f'(x)= (limit as h->0) [f(x+h)-f(x)]/(h)

The Attempt at a Solution

Here's my work:

f'(x)= (lim h->0) [(x+h)/(x+h+1) - x/(x-1)] / h

=(lim h->0) [(x+h)/(x+h+1) - x/(x-1)]/h * (x+h+1)(x+1)

= (lim h->0) [(x+h)(x+1) - (x)(x+h+1)] / (h)(x+h+1)(x+1)

= (lim h->0) (x^2+x+hx+h-x^2-xh-x)/(xh+h^2x+hx^2+h+h^2+hx)

= (lim h->0) h/(h)(x+hx+x^2+h+x)

=(lim h->0) 1/x^2+2x+hx+h

f'(x)= 1/x^2+2x

Power Rule:

y'= (x)/(x+1)
y'= (x)*(x+1)^-1
y'= -(x+1)^-2
y'= -1/(x+1)^2

Thank you so much!

 

[SammyS]

b]1. Homework Statement [/b]
Ok, so the question wants me to differentiate f(x)= (x)/(x+1). We are supposed to use the definition of the derivative f'(x)= (limit as h->0) [f(x+h)-f(x)]/(h). We also have learned the power rule. I did the formula and had some issues, so I tried to check my answer with the power rule. However they didn't come out the same. I'm not sure where I made the mistake, the formula or the power rule!

2. Homework Equations [/b
f'(x)= (limit as h->0) [f(x+h)-f(x)]/(h)

The Attempt at a Solution

Here's my work:

f'(x)= (lim h->0) [(x+h)/(x+h+1) - x/(x-1)] / h

=(lim h->0) [(x+h)/(x+h+1) - x/(x-1)]/h * (x+h+1)(x+1)

= (lim h->0) [(x+h)(x+1) - (x)(x+h+1)] / (h)(x+h+1)(x+1)

= (lim h->0) (x^2+x+hx+h-x^2-xh-x)/(xh+h^2x+hx^2+h+h^2+hx)

= (lim h->0) h/(h)(x+hx+x^2+h+x)

=(lim h->0) 1/x^2+2x+hx+h

f'(x)= 1/x^2+2x

Power Rule:

y'= (x)/(x+1)
y'= (x)*(x+1)^-1
y'= -(x+1)^-2
y'= -1/(x+1)^2

Thank you so much!

Hello mooha. Welcome to PF !

It's pretty difficult to read line after line of fractions in the form of

[(x+h)(x+1) - (x)(x+h+1)] / (h)(x+h+1)(x+1), etc,​

especially when you don't always use proper parentheses, as in

1/x^2+2x+hx+h, which I assume you intended to be $\displaystyle \frac{1}{x^2+2x+hx+h}\ .$​

You might make you life easier by writing f(x) as

$\displaystyle f(x)=1-\frac{1}{x+1}\ .$​

 

[mooha]

Unfortunately my computer does not have an easy way to do that! Unless you are able to do that with the thread options? It would make my life a lot easier! It is hard to keep putting in those parenthesis! :) Thanks

 

[SammyS]

Unfortunately my computer does not have an easy way to do that! Unless you are able to do that with the thread options? It would make my life a lot easier! It is hard to keep putting in those parenthesis! :) Thanks

I will look at it more carefully now.

 

[SammyS]

Here's my work:

f'(x)= (lim h->0) [(x+h)/(x+h+1) - x/(x-1)] / h

=(lim h->0) [(x+h)/(x+h+1) - x/(x-1)]/h * (x+h+1)(x+1)/((x+h+1)(x+1))

= (lim h->0) [(x+h)(x+1) - (x)(x+h+1)] / ((h)(x+h+1)(x+1))

= (lim h->0) (x^2+x+hx+h-x^2-xh-x)/(xh+h^2x+hx^2+h+h^2+hx)  (There's no need to mess with the denominator! )

= (lim h->0) h/((h)(x+hx+x^2+h+x))  Even with the parentheses, this denominator is not equivalent to the one in the above line.

You don't need to expand the denominator, & apparently that's where your mistake came from.

 

[mooha]

Ok thank you! So my use of the power rule is correct?

 

[mooha]

I fixed my mistake, but the answer from the power rule does not match up with the one from the equation.
WAIT! does the power rule go to zero?? because it is (x)(x+1)^-1 and the x goes to zero? does the power rule not work at all for this kind of function?

 

[SammyS]

I fixed my mistake, but the answer from the power rule does not match up with the one from the equation.
WAIT! does the power rule go to zero?? because it is (x)(x+1)^-1 and the x goes to zero? does the power rule not work at all for this kind of function?

Yes the power rule works for (x)(x+1)^-1 ...

after you use the product rule !

Try the alternate expression for f(x): f(x) = 1 - (x+1)^-1 .

 

[Ray Vickson]

Unfortunately my computer does not have an easy way to do that! Unless you are able to do that with the thread options? It would make my life a lot easier! It is hard to keep putting in those parenthesis! :) Thanks

It is hard to read such expressions without parentheses, so I don't bother trying.

RGV