can anyone help me with these differentiation problems?

Favorite Answer

1)
let Q = ax^3 + bx^2 + cx + d
Q' = 3ax^2 + 2bx + c
Q'' = 6ax + 2b
Q''' = 6a

so Q'''(1) = 36
6a = 36
a = 6

Q''(1) = 8
6ax + 2b = 8
36 + 2b = 8
2b = -28
b = -14

Q'(1) = 3
3ax^2 + 2bx + c = 3
18 - 28 + c = 3
-10 + c = 3
c = 13

and Q(1) = 1
so ax^3 + bx^2 + cx + d = 1
or 6 - 14 + 13 + d = 1
5 + d = 1
d = -4

so Q = 6x^3 - 14x^2 + 13 x - 4

2)
y = Ax^2 + Bx + C
y' = 2Ax + B
y'' = 2A
so if y '' + y ' - 2y = x^2
2A + (2Ax + B) - 2(Ax^2 + Bx + C) = x^2
-2Ax^2 +(2A-2B)x + (2A +B - 2C) = x^2

Therefore,
-2A = 1 (since the coefficient of x^2 on the right hand side is 1)
or A = -1/2
2A - 2B = 0 (since there is no x term on the right hand side)
-1 - 2B = 0
b = -1/2
2A +B - 2C = 0 (since there is no constant term on r.h.s.)
-1 - 1/2 - 2C = 0
2C = -3/2
C = -3/4...Show more

f(x) + x^2 * (f(x))^3 = 10 f'(x) + x^2 * 3 * (f(x))^2 * f'(x) + 2x * (f(x))^3 = 0 f(a million) = 2 f'(a million) + a million^2 * 3 * (2)^2 * f'(x) + 2 * a million * (2)^3 = 0 f'(a million) * (a million + 3 * 4) + 2 * 8 = 0 f'(a million) * 13 = -sixteen f'(a million) = -sixteen/13 All I used became into the product rule. enable f(x) = y y + x^2 * y^3 = 10 y' + x^2 * 3y^2 * y' + y^3 * 2x = 0 x = a million, y = 2 y' + a million * 12 * y' + 2^3 * 2 * a million = 0 y' * (a million + 12) = -sixteen y' = -sixteen/13 See? comparable element, with out the sensible notation...Show more

y=ax^3+bx^2+cx+d


y'=3ax^2+2bx+c
y''= 6ax +2b
y'''=6a =36
a=6
36+2b=8 b= -12
18-24+c=3 c=9 a+b+c+d=1 6-12+9+d= 1 d=-2
Q(x) = 6x^3-12x^2+ 9x-2
2)y''=2A ,y'=2Ax+B y= Ax^2 +Bx +C
2A+2Ax+B -2Ax^2-2Bx-2C =x^2

-2A=1 A =-1/2 (-1-2B)= 0 B=-1/2

-1 -1/2 -2C= 0 C=-3/4...Show more