Finding f'(x) using the definition of the derivative

[bktheking]

I'm having trouble finding the derivative of this function using the definition of the derivative. The function is:


f(x)= 1/x+5

I know that I'm supposed to replace x with (x+h) and subtract f(x) then divide by h. I just can't figure out how to get a common denominator between 1/(x+h)+5 and 1/(x+h) Any help would be appreciated!

 

[pka]

I'm having trouble finding the derivative of this function using the definition of the derivative. The function is: f(x)= 1/x+5
I know that I'm supposed to replace x with (x+h) and subtract f(x) then divide by h. I just can't figure out how to get a common denominator between 1/(x+h)+5 and 1/(x+h) Any help would be appreciated!

\(\displaystyle \dfrac{\frac{1}{x+5+h}-\frac{1}{x+5}}{h}\)

Now can you do the required algebra?

 

[srmichael]

I'm having trouble finding the derivative of this function using the definition of the derivative. The function is:


f(x)= 1/x+5

I know that I'm supposed to replace x with (x+h) and subtract f(x) then divide by h. I just can't figure out how to get a common denominator between 1/(x+h)+5 and 1/(x+h) Any help would be appreciated!

First, I assume the function is \(\displaystyle f(x)=\frac{1}{x}+5\)

\(\displaystyle f(x+h)=\frac{1}{x+h}+5\)

\(\displaystyle f(x+h)-f(x)=(\frac{1}{x+h}+5)-(\frac{1}{x}+5)=\frac{1}{x+h}-\frac{1}{x}\)

Can you take it from here?

 

[bktheking]

\(\displaystyle \dfrac{\frac{1}{x+5+h}-\frac{1}{x+5}}{h}\)

Now can you do the required algebra?

No. As I sated in my question, I can't get a common denominator to subtract f(x) from f(x+h).

 

[bktheking]

First, I assume the function is \(\displaystyle f(x)=\frac{1}{x}+5\)

\(\displaystyle f(x+h)=\frac{1}{x+h}+5\)

\(\displaystyle f(x+h)-f(x)=(\frac{1}{x+h}+5)-(\frac{1}{x}+5)=\frac{1}{x+h}-\frac{1}{x}\)

Can you take it from here?

The function is actually f(x)= 1/(x+5), sorry for not being clear.

 

[srmichael]

The function is actually f(x)= 1/(x+5), sorry for not being clear.

OK, so in that case you must find a common denominator between

\(\displaystyle \frac{1}{x+h+5}\ and \ \frac{1}{x+5}\)

In this case it is just the product of each so:

\(\displaystyle \frac{x+5}{(x+h+5)(x+5)}-\frac{x+h+5}{(x+h+5)(x+5)}\)

\(\displaystyle \frac{(x+5)-(x+h+5)}{(x+h+5)(x+5)}\)

Can you finish it now?

 

[bktheking]

OK, so in that case you must find a common denominator between

\(\displaystyle \frac{1}{x+h+5}\ and \ \frac{1}{x+5}\)

In this case it is just the product of each so:

\(\displaystyle \frac{x+5}{(x+h+5)(x+5)}-\frac{x+h+5}{(x+h+5)(x+5)}\)

\(\displaystyle \frac{(x+5)-(x+h+5)}{(x+h+5)(x+5)}\)

Can you finish it now?

Thank you. So (x+5) - (x+h+5)/(x+h+5) (x+5) would just be h/(x+h+5)(x+5), correct? Then I would divide by h?

 

[srmichael]

Thank you. So (x+5) - (x+h+5)/(x+h+5) (x+5) would just be h/(x+h+5)(x+5), correct? Then I would divide by h?

Close. \(\displaystyle \frac{-h}{(x+h+5)(x+5)}\)

Then, yes, you will divide by h and then take the limit as h goes to 0.

By the way, every time you do the derivative using the definition of a derivative, all terms in the numerator MUST have a "h" so you can factor and cancel with the "h" in the denominator. That's a good way to check to make sure you didn't mess up, at least as far as f(x+h) - f(x) goes.

 

[bktheking]

Close. \(\displaystyle \frac{-h}{(x+h+5)(x+5)}\)

Then, yes, you will divide by h and then take the limit as h goes to 0.

By the way, every time you do the derivative using the definition of a derivative, all terms in the numerator MUST have a "h" so you can factor and cancel with the "h" in the denominator. That's a good way to check to make sure you didn't mess up, at least as far as f(x+h) - f(x) goes.

Ah, thank you. So f'(x) would be -1/(x+5)(x+5)? And if I wanted to solve for say, f'(-3), it would be -1/4?

 

[srmichael]

Ah, thank you. So f'(x) would be -1/(x+5)(x+5)? And if I wanted to solve for say, f'(-3), it would be -1/4?

Yeeeeup!

 

[bktheking]

Thank you very much for your help good sir.