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§3.3 Differentiation formulas
Differentiation formulas: If c is a constant, and f, g are differentiable func- tions, then
d dx (c) = 0 (1) d dx (xn) = n xn−^1 (2) (c f )′^ = c f ′^ (3) (f + g)′^ = f ′^ + g′^ (4) (f − g)′^ = f ′^ − g′^ (5) (f g)′^ = f g′^ + f ′^ g (6) ( f g
)′^ =
g f ′^ − f g′ g^2
Proof of the differentiation formulas (2). First proof : If n is positive integer, we have
xn^ − an^ = (x − a) (xn−^1 + xn−^2 a + · · · + x an−^2 + an−^1 )
If f (x) = xn, f ′(a) can be found as follows.
f ′(a) = (^) xlim→a f (x) − f (a) x − a = lim x→a xn^ − an x − a = (^) xlim→a(xn−^1 + xn−^2 a + · · · + x an−^2 + an−^1 = an−^1 + an−^2 a + · · · + a an−^2 + an−^1 = n an−^1
Second proof : We can expand (x + h)n^ using Binomial Theorem:
(x + h)n^ = xn^ + n xn−^1 h + n^ (n^ −^ 1) 2 xn−^2 h^2 + · · · + n x hn−^1 + hn
And hence
f ′(x) = lim h→ 0
(x + h)n^ − x h
= lim h→ 0
xn^ + n xn−^1 h + n^ (n 2 − 1)xn−^2 h^2 + · · · + n x hn−^1 + hn^ − xn h = (^) hlim→ 0 n xn−^1 + n (n − 1) 2 xn−^2 h + · · · + n x hn−^2 + hn−^1 = n xn−^1 1
(6). Let F (x) = f (x) g(x), then
F ′(x) = lim h→ 0
F (x + h) − F (h) h = lim h→ 0
f (x + h) g(x + h) − f (x) g(x) h = (^) hlim→ 0 f (x + h) g(x + h) − f (x + h) g(x) + f (x + h) g(x) − f (x) g(x) h = (^) hlim→ 0 f (x + h) g(x + h) − f (x + h) g(x) h + lim h→ 0
f (x + h) g(x) − f (x) g(x) h = lim h→ 0 f (x + h) lim h→ 0
g(x + h) − g(x) h
f (x + h) − f (x) h lim h→ 0 g(x) = f (x) g′(x) + f ′(x) g(x) (7). Let F (x) = f (x)/g(x), then
F ′(x) = (^) hlim→ 0 F (x + h) − F (x) h = lim h→ 0 f (x + h)/g(x + h) − f (x)/g(x) h = (^) hlim→ 0 f (x + h) g(x) − f (x) g(x + h) h g(x + h) g(x) = lim h→ 0
f (x + h) g(x) − f (x) g(x) + f (x) g(x) − f (x) g(x + h) h g(x + h) g(x)
= (^) hlim→ 0
f (x+h) g(x)−f (x) g(x) h −^
f (x) g(x+h)−f (x) g(x) h g(x + h) g(x)
=
limh→ 0 f^ (x+h)^ g(x h)− f^ (x)^ g(x)− limh→ 0 f^ (x)^ g(x+h h)−f^ (x)^ g(x) limh→ 0 g(x + h) g(x) = g(x) f ′(x) − g′(x) f (x) g(x)^2 (2). When n > 0, then
(x−n)′^ = (
xn^
)′^ =
xn^ (1)′^ − 1 (xn)′ (xn)^2
−n xn−^1 x^2 n^ = −n x−n−^1
Examples: Differentiate the functions (1). f (x) =
√ 10 x^7 (2). f^ (x) = 5^ x (^8) + 2 x (^5) + 6 (3). f (x) = x (^2) − √ 41 x^3 (4). f (x) = 32 xx−+1^1 (5). f (x) =
√x− 1 √x+1 (6).f (x) = √x (x − 1) (7). f (x) = (^) x+x cx (8). f (x) = x−^3 x^
√x √x 9. f (x) = ( (^) x^12 − (^) x^34 ) (x + 5 x^3 )
Solution. (1). f ′(x) = − 7
√ 10 x^8 (2). f ′(x) = 40 x^7 + 10 x^4 (3). f ′(x) = 2 x + (^34) x √ (^41) x 3 (4).
f ′(x) = (2 x + 1) (3 x − 1)′^ − (3 x − 1) (2 x + 1)′ (2 x + 1)^2 = 3 (2 x + 1) − 2 (3 x − 1) (2 x + 1)^2 =
(2 x + 1)^2
and the corresponding change in y is
∆y = f (x 2 ) − f (x 1 )
The difference quotient ∆y ∆x = f^ (x^2 )^ −^ f^ (x^1 ) x 2 − x 1 is called the average rate of change of y with respect to x over the interval [x 1 , x 2 ]. Rates of change occur in all the sciences. Velocity, density, current, power, and temperature gradient in physics, rate of reaction and compressibility in chemistry, rate of growth and blood velocity gradient in biology, marginal cost and marginal profit in economics, rate of heat flow in geology, rate of improvement of performance in psychology, rate of spread of a rumor in sociology–these are all special cases of a single mathematical concept, the derivative.
§3.5 Derivative of Trigonometric Functions
Derivatives of Trigonometric Function d dx (sin x) = cos x d dx (csc x) = − csc x cot x d dx (cos x) = − sin x d dx (sec x) = sec x tan x d dx (tan x) = sec x d dx = − csc^2 x Two important limits
θ^ lim→ 0
sin θ θ = 1, (^) θlim→ 0 cos θ − 1 θ
We now use a geometric argument to prove that
θlim→ 0
sin θ θ = 1
O
D
A
B
C
E
θ
Figure 1
In the Figure 1, if θ > 0, we have sin θ = |BC| < |AB| < arcAB = θ
And hence sin θ θ
On the other hand,
θ = arcAB < |AE| + |EB| < |AE| + |ED| = |AD| = |OA| tan θ
hence
θ < sin θ cos θ i.e., sin θ θ
cos θ
From
cos θ < sin θ θ
and limθ→ 0 + cos θ = 1 = limθ→ 0 + 1 = 1. Hence, we have
lim θ→ 0 +
sin θ θ = 1 But sin(−θ)/(−θ) = sin θ/θ, so we have
θlim→ 0
sin θ θ
We can now prove the limit
lim θ→ 0
cos θ − 1 θ
as follows:
lim θ→ 0
cos θ − 1 θ = lim θ→ 0
cos θ − 1 θ
cos θ + 1 cos θ + 1
= lim θ→ 0
cos^2 θ − 1 θ (cos θ + 1
= (^) θlim→ 0 − sin^2 θ θ (cos θ + 1) = − (^) θlim→ 0 sin θ θ
sin θ cos θ + 1 = − (^) θlim→ 0 sin θ θ · (^) θlim→ 0 sin θ cos θ + 1 = − 1 · (
Now, we can prove the derivatives of trigonometric functions. d dx (sin x) = (^) hlim→ 0 sin(x + h) − sin x h = lim h→ 0
sin x cos h + cos x sin h − sin x h = (^) hlim→ 0 sin x (cos h − 1) h
- lim h→ 0 cos x sin h h = sin x (^) hlim→ 0 cos h − 1 h
- cos x (^) hlim→ 0 sin h h = sin x (0) + cos(x) (0) = cos x Prove (cos x)′^ = − sin x by yourself.
Solution.
x^ lim→ 0
sin 7 x 4 x = (^) xlim→ 0 sin 7 x 7 x
7 lim x→ 0
sin 7 x 7 x =
Example 4 Find
xlim→ 0 x^ cot^ x
Solution.
x^ lim→ 0 x^ cot^ x^ =^ xlim→ 0
x cot x sin x = (^) xlim→ 0 cos x sin x x
limx→ 0 cos x limx→ 0 sinx^ x = cos 0 1 = 1 §
Assignment
- Differentiate
(a). f (x) = a x + b c x + d (b).^ f^ (x) =^
x^3 + x x^4 − 2 (c). f (x) = x^3 cos x (d). f (x) = sec x 1 + sec x (e). f (x) = 1 + sin^ x x + cos x (f ). f (x) = x sin x cos x
- Find the limit
(a). (^) xlim→ 0 sin 4 x sin 6 x (b). lim x→π/ 4
sin x − cos x cos 2 x (c). lim t→ 0 tan 6 t sin 2 t (d).^ xlim→ 1
sin(x − 1) x^2 + x − 2
- If f (3) = 4, g(3) = 2, f ′(3) = −6, and g′(3) = 5, find the following numbers.
(a). (f g)′(3), (b).( f f − g
- Let f (x) =
x^2 if x ≤ 2 m x + b if x > 2 Find the values of m and b that make f differentiable everywhere.
Exercise
The exercises are not graded. Completing these exercises is left to the
discretion of the student.
1. Read section 3.4 by yourself.
2. Page 155, 23-42; 51-54;
3. Page 174: 1-16; Page 175: 35-
