MATH241 Lecture Notes - Lecture 10: Differentiable Function
MATH241 - Lecture 10 - Derivatives and Differentiable Functions
2.8: The Derivative as a Function
Example
:
Determine the derivative
A) (x)f=x3+ 2x2− 8
B) (x)g=x
x + 1
C) (x)h=x2
3=√x3
Solution:
(x)f′= lim
h → 0 h
f(x + h) − f(x)
A) (x)f′= lim
h → 0 h
(x + h) + 2(x + h) − 8 − x − 2x + 8
3 2 32
= lim
h → 0 h
x + 3x h + 3x h + h + 2x + 4xh + 2h − x − 2x
32 2 3 2232
= lim
h → 0 h
3x h + 3xh + h + 4xh + 2h
2 2 3 2
= lim
h → 0 h
3x + 3xh + h + 4x + 2h
2 2
x= 3x2+ 4
B) (x)g′= lim
h → 0 h
−
x + h
x + h + 1 x
x + 1
= lim
h → 0 (h
1)[(x + h + 1) (x+ 1)
(x + h) (x + 1) − x (x + h + 1) ]
= lim
h → 0 (h
1)[(x + h + 1) (x + 1)
x + x + hx + h − x − xh − x
2 2 ]
= lim
h → 0 (h
1)[h
(x + h + 1) (x + 1) ]
= lim
h → 0
1
(x + h + 1) (x + 1)
=1
(x + 1)2
C) (x)h′= lim
h → 0 h
−
√(x + h)3√x3
= lim
h → 0 h
−
√(x + h)3√x3· +
√(x + h)3√x3
−
√(x + h)3√x3
Document Summary
Math241 - lecture 10 - derivatives and differentiable functions. 3 x + 3x h + 3x h + h + 2x + 4xh + 2h x 2x. 3x h + 3xh + h + 4xh + 2h. 3x + 3xh + h + 4x + 2h. B) (x + h + 1) (x+ 1) (x + h) (x + 1) x (x + h + 1)] 2 x + x + hx + h x xh x (x + h + 1) (x + 1) 2 (x + h + 1) (x + 1)] h x x + 1 x + h x + h + 1 h (x)g . = lim (x + h + 1) (x + 1) h 0 (x + 1)2. 3 x + 3x h + 3xh + h x (x + h) x3. Let f (x) y = f (x) dy. All indicate the operation of differentiation dy = lim.