Second Derivative--Calculus?

Question

I have a calculus problem that involves the chain rule (note: I am in Calculus I with Analytical Geometry).

Find the Second Derivative (y") of

(1+ (1/x)) to the power of three.

The equation is (1+(1/x)) with the entire equation to the power of 3.

(1+(1/x))^3

Answer 1

f (x) = (1 + 1/x)³

f `(x)
= 3.(1 + 1/x)².(- 1/x²)
= (- 3/x²).(1 + 1/x)²

f "(x)
= (6/x³).(1 + 1/x)² + 2.(1 + 1/x).(-1/x²).(- 3/x²)
= (6/x³).(1 + 1/x)² + (6 / x^4).(1 + 1/x)
= (6 /x³).(1 + 1/x).[ (1 + 1/x) + 1/x ]
= (6/x³).(1 + 1/x).(1 + 2/x)

Answer 2

It may help to do a u substitution.

y = (1+1/x)^3 Let u = 1 + 1/x

Then y' = f'(u)*u'

y = u^3
y' = 3*u^2 * u'
y' = 3* (1 + 1/x)^2 * (-1/x^2)

For y'' you will also need the product rule.

y'' = 3*(1+1/x)^2 * (2/x^3) + (-1/x^2)*derivative ( 3*(1+1/x)^2)

Using u substitution again:

-------------A-------
y'' = [6*(1+1/x)^2]/x^3 -1/x^2 * 6u*u'
= A -1/x^2* 6*(1+1/x)*-1/x^2
= A + [6*(1+1/x)] / x^4

Answer 3

y = (1+ (1/x))³

y' = 3 (1+ (1/x))² · (-x^-2) = -3(1+ (1/x))² · (x^-2)
next derivative also needs product rule...

y'' = 6(1+ (1/x)) · (x^-2)² - 3 (1+ (1/x))² · (-2x^-3)
= 6(1+ (1/x)) · (x^-2)² + 6 (1+ (1/x))² · (x^-3)


d:

Answer 4

y = (1+ (1/x))^3
= 1+ 3/x + 3/x^2 + 1/x^3
y" = 6/x^3 + 18/x^4 + 6/x^5

Answer 5

y=(1+(1/x))^3
First Derivative
y'= 3(1+(1/x))^2 times (-x)^-2

Second Derivative
the product rule: f'g+g'f
y"= 6(1+(1/x))(-x^-2)(-x^-2) + (2)(x)^-3 times (3(1+(1/X))^2

Answer 6

(6(x+1)(x+2))/(x^5)

Answer 7

ok
First note that the equation can be rewriten into:
y=1+x^(-3)
Take the first deriv:
y'=-3x^-4
Take the second deriv:
y''=12x^-5
Which is also the same as :
y''=12(1/x)^(5)