By the definition of the derivative, we have
( uv )′( x ) = lim Δy Δu ⎞ Δv ⎞ . ⎛ ⎛ lim ( ) = lim ⎜ v ( x + Δx ) + u x ⎜ Δx → 0 Δ x Δx → 0 Δx ⎟ Δx ⎟ ⎝ ⎠ Δx → 0 ⎝ ⎠
(
)
y=
( x − 1) ( x 2 − 2 x )
x4
x 3 − 3 x 2 + 2 x . −1 −2 −3 = = x − 3 x + 2 x . 4 x
Then use the Sum and Power Rules:
dy 1 6 6 . 2+ 3− 4. = − x − 2 − 3 ( − 2 ) x − 3 + 2 ( − 3 ) x −4 = − dx x x x
Derivation rules for sum, difference, product and quotient of functions
1 2 1 ( x + ). . x x 1 x2 + 1 . Solution: By rules (2) with u = and v =. : x x
Since v ( x ) is derivable at x , it must be continuous at x and hence Therefore Finish.
3
Δx → 0
lim v ( x + Δx ) = v ( x ) .
( uv )′( x ) = u′( x )v ( x ) + u( x )v ′( x ) .
4
Derivation rules for sum, difference, product and quotient of functions
Example: Find the derivative of y = x 2 − 2 x + 3cos x + x ln x . Solution: By rules (1) and (2), we have
dy d d 1⎞ 2 ⎛ 1 ⎞ = = + − = − 1 2 1 . ( x) + 2 ⎛ . ⎜ ⎟ ⎜ 2 ⎟ 2 dx dx dx ⎝ x ⎠ x ⎝ x ⎠
The slope at x = 1 is
dy dx 2⎤ ⎡ 1 = − x =1 ⎢ x2 ⎥ ⎣ ⎦
x =1
= 1 − 2.= −1.
The line through (1,3) with slope m = -1 is
y − 3 = ( −1)( x − 1) y = −x +1+ 3 y = − x + 4.
9
.
(cot x )′ = − csc x
2
(tan x )′ = sec 2 x
(cot x )′ = − csc 2 x
6
Derivation rules for sum, difference, product and quotient of functions
t2 − 1 . Example(P103): Find the derivative of y = 2 . t +1 Solution: We apply the Quotient Rule with u = t 2 − 1 and v. = t 2 + 1,
Finish.
ln x 2 x
+
1 . x
5
Derivation rules for sum, difference, product and quotient of functions
Example: Find the derivative of y = tan x and y = cot x . Solution: By the quotient rule, we have
⎛ sin x ⎞′ (sin x )′ cos x − (cos x )′ sin x (tan x )′ = ⎜ ⎟ = ⎝ cos x ⎠ cos 2 x
That is
cos 2 x + sin 2 x 2 = = sec 2 cos x
(tan x )′ = sec 2 x
By the same way, we have Finish.
dy ( t = dt = =
2
.
+ 1 i 2t − t 2 − 1 i 2t
2 2
) ( ( t + 1) )
2
)
2t 3 + 2t − 2t 3 + 2t
(t
2
+1
2
.
(t
4t
2
+1
)
.
7
Derivation rules for sum, difference, product and quotient of functions
Find the derivative of y =
d ⎡ 1 ⎛ 2 1 ⎞⎤ 1 ⎛ 1 ⎞ ⎛ 2 1 ⎞⎛ 1 ⎞ ⎜ x + x ⎟ ⎥ = x ⎜ 2 x − x 2 ⎟ + ⎜ x + x ⎟⎜ − x 2 ⎟ dx ⎢ x ⎠⎦ ⎝ ⎠ ⎝ . ⎠⎝ ⎠ ⎣ ⎝ 1 1 2 = 2 − 3 −1− 3 = 1− 3 . Finish. x x x
8
Derivation rules for sum, difference, product and quotient of functions
Example(P103): Find an equation for the tangent to the curve 2 . at the point (1,3). y= x+ x Solution: The slope of the curve is
dy = ( x 2 )′ − (2 x )′ + (3cos x )′ + ( x ln x )′ dx = 2 x − 2 x ln 2 − 3sin x + ( x )′ ln x + x (ln x )′
= 2 x − 2 x ln 2 − 3sin x + ln x 2 x + x x
= 2 x − 2 x ln 2 − 3sin x +
Section 2.2
Fundamental Derivation Rules
1
Derivation rules for sum, difference, product and quotient of functions
Theorem: (Derivation rules of rational operations) Suppose that the function u , v : I → R are derivable at x ∈ I ; then their sum, difference, product and quotient are all derivable at x , and (1) ( u ± v )′( x ) = u′( x ) ± v ′( x ) (2) ( uv )′( x ) = u′( x )v ( x ) + u( x )v ′( x ) u′( x )v ( x ) − u( x )v ′( x ) ⎛ u ⎞′ (3) ⎜ ⎟ ( x ) = (v ≠ 0) 2 v ( x) ⎝v⎠ In particular u′( x ) ⎛ 1 ⎞′ ( u( x ) ≠ 0) . (cu)′( x ) = cu′( x ) ( c ∈ R is a constant), ⎜ ⎟ ( x ) = − 2 u ( x) ⎝ u⎠
Example(P103): Choosing which Rule to use ,rather than using Quotient Rule ( x − 1) x 2 − 2 x . . to find the derivative of y = 4 x . Solution: Expand the numerator and divide by x 4 :
2
Derivation rules for sum, difference, product and quotient of functions
Proof: Let’s prove only the rule (2). Let y = u( x )v ( x ) , then
Δy = u( x + Δx )v ( x + Δx ) − u( x )v ( x ) = u( x + Δx )v ( x + Δx ) − u( x )v ( x + Δx ) + u( x )v ( x + Δx ) − u( x )v ( x ) = v ( x + Δx )Δu + u( x )Δv .