Finding the derivative with first principle

f(x) = sqrt (4-x²)

Can someone show me how to find the derivative using lim (h-->0)? Much appreciated, thanks!

√(4 - (x + h)²) - √(4 - x²)
------------------------------ =
···············h

√(4 - (x + h)²) - √(4 - x²) √(4 - (x + h)²) + √(4 - x²)
------------------------------- -------------------------------- =
·············· h ·············· √(4 - (x + h)²) + √(4 - x²)

4 - (x + h)² - (4 - x²)
------------------------------------ =
h (√(4 - (x + h)²) + √(4 - x²))

·········x² - (x + h)²
------------------------------------- =
h (√(4 - (x + h)²) - √(4 - x²))

········· -2xh - h²
------------------------------------ =
h (√(4 - (x + h)²) + √(4 - x²))

········· -2x - h
------------------------------- =
√(4 - (x + h)²) + √(4 - x²)

limit as h --> 0 is

···-x
-----------
√(4 - x²)

f(x) = sqrt (4-x²)
Let's re-write this as f(x)=(4-x^2)^1/2
Now we make use of the chain rule, which is the derivative of the "outside"term times the derivative of the "inside" term.
Outside derivative = 1/2(4-x^2)
Inside = -2x
multiply together to get f ' (x)= 1/2(4-x^2) (-2x)

It is not necessary to use the definition... just use derivative rules...

First write it so f(x) = (4-x^2)^(1/2) ... use chain rule f´(x) = -2x*(1/2)(4-x^2)^(-1/2)

then f´(x) = -x/ sqrt(4-x^2) OK!

[rt(4-(x+h)^2) - rt(4-x^2)] [rt(4-(x+h)^2) + rt(4-x^2)]/h[rt(4-(x+h)^2) + rt(4-x^2)] = [4-(x+h)^2 -(4-x^2)]/h[rt(4-(x+h)^2 + rt(4-x^2)] = [-2xh - h^2]/h(...) = (-2x-h)/[rt(4-(x+h)^2) + rt(4-x^2)] , f' = (-2x)/2rt(4-x^2) = -x/rt(4-x^2)

http://www.math.montana.edu/frankw/ccp/c…

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