how do you find the second derivative of...?

Question

(x+3)/(x-1)

Answer 1

f (x) = (x + 3) / (x - 1)
Using quotient rule : -
f `(x) = [ (x - 1) - (x + 3) ] / (x - 1) ²
f `(x) = - 4 / (x - 1) ²
f "(x) = 4 (2)(x - 1) / (x - 1) ^ 4
f "(x) = 8 / (x - 1) ³

Answer 2

First derivative: (1/x-1) - ((x+3)/(x-1)^2)

Second derivative: (2(x+3))/(x-1)^3 - 1/(x-1)^2 - 1/(x^2)

Answer 3

y = (x+3)/(x-1)
= (x-1+4)/(x-1)
= 1 + 4/(x-1)
= 1 + 4(x-1)^-1

dy/dx = 0 - 4(x-1)^-2
= -4(x-1)^-2

d²y/dx² = 8(x-1)^-3
= 8/(x-1)³

Answer 4

(x+3)/(x-1)

Let u = (x+3)
Let v = (x-1)

d/dx (x+3)/(x-1) = (vu' - uv')/v² or (v * d/dx(u) - u * d/dx(v))/v²

= (x-1) d/dx(x+3) - (x+3) d/dx(x-1))/(x-1)² =
(1 *(x-1) - 1 * (x+3)) / (x-1)²
(x-1 -x - 3)/(x-1)²
= (-4/(x-1)²)

second derivate

-4/(x-1)² = -4 * (x-1) ^ -2

d/dx (-4 * (x-1) ^ -2) = -4 * -2 (x-1) ^ -3 or 8(x-1)^-3
~~~

Answer 5

f(x) = (x+3) / (x-1)
f'(x) = [(x-1)*1 - (x+3)*1] / (x-1)^2
= -4 / (x-1)^2
f"(x) =[ (x-1)^2*0 + 4*2 (x-1) ] / (x-1)^4
= 8 / (x-1)^3

Answer 6

use the quotient rule:

d(u/v)/d(x) = (v*d(u)/d(x)-u*d(v)/d(x))/(v²)

where in this case:

u = x+3
v=x-1

d(y)/d(x) = ((x-1)-(x+3))/(x-1)²
= (-4)/(x-1)² = -4/(x-1)² = -4/x²-2x+1

then proceed to continue and find d²y/dx²

d²y/dx² = (((x²-2x+1)*0)-((-4)(2x-2)))/((x-1)^4)
= (4(2x-2))/((x-1)^4) = (8x-8)/((x-1)^4)
= (8(x-1))/((x-1)^4) = (8)/((x-1)³)

Answer 7

first find the first derivative and after that, the second derivative

f'' = (f')'

Answer 8

d/dx of f(x)/g(x) is (f(x)g'(x) -g(x)f'(x))/ (g(x))^2

-4/(x-1)^2