SOLUTION: Finding the derivative using the defintion? hi i would like to find out how to solve this problem. i would like to find the derivative of the function using the definition of th

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Finding the derivative using the defintion? hi i would like to find out how to solve this problem. i would like to find the derivative of the function using the definition of th      Log On


   

Question 79723: Finding the derivative using the defintion?
hi i would like to find out how to solve this problem. i would like to find the derivative of the function using the definition of the derivative and state the domain of the function and the domain of its derivative
g(x) = 1/x^2
thanks for any help. it is very much appreciated
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Using the definition of the derivative:
f'(x)=%28f%28x%2Bh%29-f%28x%29%29%2Fh%29

%281%2F%28x%2Bh%29%5E2-1%2Fx%5E2%29%2Fh Start with this expression by plugging in (x+h) into x of 1%2Fx%5E2

Multiply both fractions to get a common denominator

Multiply

%28%28x%5E2-%28x%2Bh%29%5E2%29%2F%28x%5E2%28x%2Bh%29%5E2%29%29%2Fh Combine the fractions

%28%28x%5E2-%28x%5E2%2B2xh%2Bh%5E2%29%29%2F%28x%5E2%28x%2Bh%29%5E2%29%29%2Fh Foil %28x%2Bh%29%5E2

%28%28x%5E2-x%5E2-2xh-h%5E2%29%2F%28x%5E2%28x%2Bh%29%5E2%29%29%2Fh Distribute the negative

%28%28-2xh-h%5E2%29%2F%28x%5E2%28x%2Bh%29%5E2%29%29%2Fh Combine like terms

%28h%28-2x-h%29%2F%28x%5E2%28x%2Bh%29%5E2%29%29%2Fh Factor out an h

%28h%28-2x-h%29%2F%28h%28x%5E2%28x%2Bh%29%5E2%29%29%29 Multiply the first fraction by the reciprocal of the 2nd

%28cross%28h%29%28-2x-h%29%2F%28cross%28h%29%28x%5E2%28x%2Bh%29%5E2%29%29%29 Notice these h terms cancel

So we're left with this:

%28%28-2x-h%29%2F%28%28x%5E2%28x%2Bh%29%5E2%29%29%29

%28%28-2x-h%29%2F%28%28x%5E2%28x%5E2%2B2xh%2Bh%5E2%29%29%29%29 Foil the denominator

%28%28-2x-h%29%2F%28%28x%5E4%2B2x%5E3h%2Bx%5E2h%5E2%29%29%29%29 Distribute the x%5E2

%28%28-2x-%280%29%29%2F%28%28x%5E4%2B2x%5E3%280%29%2Bx%5E2%280%29%5E2%29%29%29%29 Now plug in h=0 since we're taking a limit to zero

So it all simplifies to this:

%28-2x%29%2F%28x%5E4%29

Now divide to simplify

%28-2%29%2F%28x%5E3%29

So according to the definition, the derivative of 1%2Fx%5E2 is


%28-2%29%2F%28x%5E3%29