Expert: Scotto - 1/18/2010

Question
I am working on a problem to find the derivative of
f'(x) if f(x) = 3/x^2 + 2/ sqrt x

and we have to use a 4 step process..
example
Step 1: find f(x+h)
Step 2: find f (x+h) - f(x)
Step 3: find f (x+h) - f(x)/h
Step 4: find lim H -> O f  (x+h) - f(x)/h

Answer
Step 1: If f(x) = 3/x² + 2/√x, then f(x+h) = 3/(x+h)² + 2/√(x+h).

Step 2: If we take f(x+h) - f(x), we get 3/(x+h)² + 2/√(x+h) - 3/x² - 2/√x.
Coombine the 1st and 3rd and also combine the 2nd and 4th.
This gives (3x² - 3(x+h)²)/(x²(x+h)²) - 2(√x - √(x+h))/(√x√(x+h)).

Step 3: Multiply out the numerators in the 1st and 2nd fractions.

In the 1st combined fraction, the x² terms will cancel.
This leaves us with (-6xh - 3h²)/(x²(x+h)²).
When this difference is divided by h and a -3 is factored out in the numerator,
what is left is -3(2x + h)/(x²(x+h)²).

In the 2nd combined fraction, multiply the top and bottom by (√x + √(x+h)).
This is known as the conjugate of (√x - √(x+h)).
This should give you 2(x - x - h)/((√x√(x+h)2(√x + √(x+h)).

In the numerator, there is a x - x, and that cancels.
That gives us -2h/((√x√(x+h)2(√x + √(x+h)).

Now when we finally divide by h, we get -2//((√x√(x+h)2(√x + √(x+h)).

4. Letting h go to 0 in the first fraction gives us -3(2x)/(x²x²).
This simplifies to -6/x^3.

Letting h go to 0 gives us -2/((√x√x)2(2√x)).  
One of 2's in the top and bottom cancel, leaving -1/(2x√x).