Taking the derivative of a polynomial fraction?

In summary, the question involves differentiating a polynomial fraction using the definition of the derivative and the power rule. The student attempted to solve the problem but encountered issues and decided to check their answer using the power rule. However, the two answers did not match and they are unsure where they made a mistake. The expert suggests using proper parentheses in the expressions and fixing a mistake in the denominator. The power rule can be used for this type of function, but the product rule must be applied first. The expert also suggests using an alternate expression for the function to make it easier to differentiate.
  • #1
mooha
5
0
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!
 
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  • #2


mooha said:
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 [itex]\displaystyle \frac{1}{x^2+2x+hx+h}\ .[/itex]​

You might make you life easier by writing f(x) as
[itex]\displaystyle f(x)=1-\frac{1}{x+1}\ .[/itex]​
 
  • #3


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
 
  • #4


mooha said:
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.
 
  • #5


mooha said:
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.
 
  • #6


Ok thank you! So my use of the power rule is correct?
 
  • #7


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?
 
  • #8


mooha said:
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 .
 
  • #9


mooha said:
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
 

What is the definition of a derivative?

A derivative is a mathematical concept that represents the rate of change of a function at a specific point. It is the slope of the line tangent to the function at that point.

Why do we need to take the derivative of a polynomial fraction?

Taking the derivative of a polynomial fraction allows us to find the slope of the tangent line at any point on the function. This can help us analyze the behavior of the function and make predictions about its behavior in the future.

What is the process for taking the derivative of a polynomial fraction?

The process for taking the derivative of a polynomial fraction involves using the power rule, which states that the derivative of x^n is nx^(n-1). We also need to apply the chain rule if the polynomial fraction contains other functions, such as trigonometric functions or logarithms.

What are the common mistakes when taking the derivative of a polynomial fraction?

Some common mistakes when taking the derivative of a polynomial fraction include forgetting to apply the chain rule, not simplifying the resulting expression, and incorrect use of the power rule. It is important to carefully follow the steps and double check for any mistakes.

How can I check if I have correctly taken the derivative of a polynomial fraction?

You can check your work by graphing the original function and the derivative function on a graphing calculator. The derivative function should have the same slope as the tangent line at any given point on the original function. Additionally, you can also plug in specific values for x to see if the resulting values match the slope of the tangent line at those points.

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