..READ THIS
FIRST..Problems
about Differentiation of functions
READ THIS
FIRST
If a problem is solved. It is not 'the' answer.No attempt is made to search for the most elegant answer.
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Problems about Differentiation of functions
Level 1 problems
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
Level 2 problems
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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 x2With 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 - 2First 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
Level 3 problems
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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 -> inftysin(3/x) = lim ----------- x -> infty (1/x) Let t = 1/x sin(3t) = lim ----------- 0 t 3.cos(3t) = lim ----------- = 3 0 1
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