y = 5x ; y' = 5
y = x3 ; y' = 3x2
y = (2x2+ x + 5) ; y'= 4x + 1
x - 1 (x+1).1 - (x-1).1 2
y = ------- ; y' = -------------------- = ------------------
x + 1 (x+1)(x+1) (x+1)(x+1)
y = (x+5)6 ; y' = 6.(x+5)5
y = (2x + 6)3 ; y' = 3.(2x + 6)2.2
y = (2x + 6)(5x - 7) ; y'= 2.(5x - 7) + (2x + 6).5
y = (2x + 6)2.(5x - 7)3; y'= 2.(2x + 6).2.(5x - 7)3+ (2x + 6)2.3.(5x - 7)2.5
y = x/5 ; y' = 1/5
1
y = 5/x ; y' = -5.---
x.x
7
y = --- <=> y = 7.x-4 ; y' = -28 x-5
x4
Calculate the derivative of
ax + b
y = ---------
cx + d
|
(cx + d)a - (ax + b)c ad - bc
y' = --------------------- = ----------------
(cx + d)(cx + d) (cx + d)(cx + d)
Calculate the derivative of
ad - bc
y = ----------------
(cx + d)(cx + d)
|
Since (ad -bc) is constant we write y as
y = (ad -bc). (cx + d)-2
Then
y' = (ad -bc).(-2) (cx + d) -3 .c
So,
-2c(ad -bc)
y' = -----------------
(cx + d)3
Calculate the first and second derivative of
x.x + x + 1
y = -------------
x + 1
|
(2x+1)(x+1)-(xx+x+1) x(x+2)
y' = -------------------- = ------------
(x+1)(x+1) (x+1)(x+1)
(2x+2)(x+1)2-2(x+1)(x2+2x) 2
y" = --------------------------- = ---------
(x+1)4 (x+1)3
Calculate the first and second derivative of
y = sqrt(x3-3x +2)
|
3x2-3 3(x2-1)
y' = ----------------- = ----------------------
2.sqrt(x3-3x +2) 2.sqrt( (x-1)2(x-2) )
It is difficult to calculate the second derivative from this result.
Therefore we simplify the result in the following way.
For all x > 1
3 (x +1)
y' = ----.-----------
2 sqrt(x-2)
Then
3 sqrt(x-2) -(x +1).1/(2.sqrt(x-2) )
y" = ---. ---------------------------------- = ...
2 (x-2)
3 (x + 3)
y" = ---.---------------
4 (x-2)sqrt(x-2)
For all x < 1
3 (x +1)
y' = - ---.-----------
2 sqrt(x-2)
Then
3 sqrt(x-2) -(x +1).1/(2.sqrt(x-2) )
y" = - ---. ---------------------------------- = ...
2 (x-2)
3 (x + 3)
y" = - ---.----------------
4 (x-2)sqrt(x-2)
Calculate the first and second derivative of
y = sin(2x) - cos(2x)
|
y' = 2cos(2x) + 2sin(2x)
y" = -4sin(2x) + 4cos(2x)
An area A depends on x in the following way.
2r3.x3
A = -------------- with r = constant.
(x2 + r2)2
Calculate A'.
|
(x2 + r2)2.3x2- x3.2(x2 + r2).2x
A' = 2r3 . -------------------------------------
(x2 + r2)4
= ...
x2.(3r2 - t2)
A' = 2r3.-----------------
(x2 + r2)3
Find
1 - sqrt(cos(x))
lim ------------------
0 x2
|
With l'Hospital's rule
sin(x)/(2.sqrt(cos(x)))
= lim ------------------------
0 2x
sin(x) 1
= lim -------.------------
0 x 4sqrt(cos(x))
= 1.(1/4) = 1/4
Find
x(2x + 3)x - 1
lim -------------------
-1 x3 - 3x - 2
|
First method:
(x+1)(x+1)(2x-1)
= lim ---------------
-1 (x+1)(x+1)(x-2)
(2x-1)
= lim -------- = 1
-1 (x-2)
Second method:
2x3 + 3x2 -1
lim -------------------
-1 x3 - 3x - 2
With l'Hospital's rule
6x2 + 6x
= lim -------------
-1 3x2 - 3
With l'Hospital's rule
12x + 6
= lim ------------- = 1
-1 6x
x2 + 2px + q
Given : y = --------------
x2 + 1
Prove that there are two x values, x' and x", such that y' = 0.
Then prove that x'.x" = -1.
|
(x2 + 1)(2x + 2p) - (x2 + 2px + q).2x
y' = --------------------------------------
(x2 + 1)2
= ...
-2px2 + 2(1 - q)x +2p
= -------------------------
(x2 + 1)2
y' = 0 if and only if -2px2 + 2(1 - q)x +2p = 0
This gives two values x' and x".
It is immediate that x'.x" = -1
The derivative of f(x) is f'(x).
The derivative of f'(x) is f"(x) an is called the second derivative of f(x).
The derivative of f"(x) is f"'(x) an is called the third derivative of f(x).
...
Now, let f(x) = sqrt(x)
n-1 (2n-2)! (1-2n)/2
Prove that the n-th derivative of f(x) = (-1) .--------.(4x)
(n-1)!
|
f(x) = x1/2
f'(x) = (1/2). x(-1/2)
f"(x) =(1/2)(-1/2) . x(-1/2-1)
f"'(x) =(1/2)(-1/2)(-1/2 - 1) . x(-1/2-2)
f""(x) =(1/2)(-1/2)(-1/2 - 1)(-1/2 - 2) x(-1/2-3)
...
So, the n-th derivative of f(x)
=(1/2)(-1/2)(-1/2 - 1)(-1/2 - 2)...(-1/2 -(n-2)) x(-1/2-(n-1))
n-1 1
= (-1) .----.(1+2).(1+2.2)(1+3.2)...(1+(n-2).2). x(-1/2-(n-1))
2n
n-1 1
= (-1) .----.3.5.7.9...(2n-3). x(1-2n)/2
2n
n-1 1 1.2.3.4.5.6....(2n-3)(2n-2)
= (-1) .----.---------------------------- x(1-2n)/2
2n 1.2.4.6. ... (2n-2)
n-1 1 1.2.3.4.5.6....(2n-3)(2n-2)
= (-1) .----.---------------------------- x(1-2n)/2
2n 2n-1. 1.2.3... (n-1)
n-1 (2n-2)!
= (-1) .-------------------- x(1-2n)/2
22n-1 .(n-1)!
n-1 (2n-2)!
= (-1) .------------.2 (1-2n) x(1-2n)/2
(n-1)!
n-1 (2n-2)!
= (-1) .------------.(4x)(1-2n)/2
(n-1)!
Find
lim (x. sin(3/x))
x -> infty
|
sin(3/x)
= lim -----------
x -> infty (1/x)
Let t = 1/x
sin(3t)
= lim -----------
0 t
3.cos(3t)
= lim ----------- = 3
0 1