Examples
involving Derivatives
|
Find the derivative
of f(x) =
1/(2x + 1) using the definition
of the derivative df(x)/dx = lim [ f(x + h)
- f(x)]/ h h → 0 f(x + h) =
1/(2x + 2h + 1), f(x) =
1/(2x + 1) f(x + h) – f(x) =
1/(2x + 2h + 1) - 1/(2x + 1) ;
put over a common denominator = [2x + 1 - (2x + 2h + 1) ] / [(2x + 2h +
1)(2x + 1)] = [ -2h ] / [(2x + 2h + 1)(2x + 1)] Now [f(x + h) – f(x)] / h
= -2 / [(2x + 2h + 1)(2x + 1)] So df/dx = lim [ f(x + h)
- f(x)]/ h =
-2/(2x + 1)2 which
is the result. h → 0 |
|
Other Examples: f(x)
= x5 df/dx = 5 x4 f(x)
= 3 x -5 ,
df/dx =
(3)(-5) x -6 = -15 x -6 f(x)
= x ( x3 +
3x ) Use product rule for this
example. df/dx = (1)( x3 + 3x
) + x ( 3x2 +
3) = 4x3 +
6x |
|
Let
f(x) = ( 6x4 + 3x
) / (x – 1) = ( 6x4 + 3x
) (x – 1)-1 Changed from quotient to product. Then use product rule. df/dx = ( 24 x3 + 3 ) (x – 1)-1 + ( 6x4 + 3x
) (-1) (x – 1)-2 Put over a common
denominator (algebraic manipulation only) df/dx = [( 24 x3 + 3 ) (x – 1) + ( 6x4 + 3x
) (-1)] / (x – 1)2 du/dx = [
18 x4 - 24 x3 - 3
] / (x – 1)2 |
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