How Do You Calculate the Derivative of 1/(x-1) Using the Definition?

[TheKracken]

Homework Statement

Find f'(x) of f(x)= 1 over x-1
using the definition of a derivative

Homework Equations

definition of a derivative is
f'(x)= lim as h→ 0 of f(x+h) - f(X) ALL OVER h

The Attempt at a Solution

I have no idea how to do this using the definition of a derivative...

 

[SammyS]

Homework Statement

Find f'(x) of f(x)= 1 over x-1
using the definition of a derivative

Homework Equations

definition of a derivative is
f'(x)= lim as h→ 0 of f(x+h) - f(X) ALL OVER h

The Attempt at a Solution

I have no idea how to do this using the definition of a derivative...

If \displaystyle f(x)=\frac{1}{x-1}\,, then what is f(x+h) ?

 

[TheKracken]

If \displaystyle f(x)=\frac{1}{x-1}\,, then what is f(x+h) ?

This is where I draw a blank...

is it simpily (1/x-1 +h) - f(x) over h meaning it cancels out to be h/h or 0?

 

[Mark44]

No.
All Sammy asked you was what is f(x + h)?

In the right side of the formula equation, replace x by x + h. That's how function notation works.

 

[SammyS]

This is where I draw a blank...

is it simply (1/x-1 +h) - f(x) over h meaning it cancels out to be h/h or 0?

Don't forget that f(x) is 1/(x-1), and No, h does not cancel the way you are supposing it does.

You will have:

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

You need to add the two fractions in the numerator -- use a common denominator.

After some simplification, you will get the h in the overall denominator to cancel.

Now we see why algebra skills are important for Calculus.