What is f ' ' ( 1/x ) ?

Question

Show your work. I'm trying to disprove a professor...

Answer 1

are you asking for the second derivative of (1/x), then

((1' * x) - (x' * 1))/((x)^2) = (0 - (1))/(x^2) = (-1/(x^2))

((-1' * x^2) - (-1 * x^2'))/((x^2)^2) = (0 - (-1 * 2x))/(x^4) = (2x)/(x^4) = 2/(x^3)

Answer 2

2 / x^3

Answer 3

I assume you mean f''(x) if f(x)=1/x,
otherwise f(1/x) or f'(1/x) or f''(1/x) is meaningless
w/o an "=" sign and the right side of the equation.

With that put 1/x in the form of x^(-1), then

f'(x)= -x^(-2) and f''(x)=2x^(-3)=2/x^(3)

Answer 4

Given the function
f(x) = 1/x

we can rewrite 1/x = x‾¹
f(x) = x‾¹

We can directly use the differentiation technique f'(xⁿ) = nxⁿ ‾ ¹
f'(x) = (-1)x‾¹ ‾ ¹= -x‾²

We can differentiate again using the same technique f'(xⁿ) = nxⁿ ‾ ¹
f"(x) = -(-2) x‾² ‾ ¹ = 2x‾³

We can now rewrite x‾ⁿ = 1/xⁿ. Therefore,
f"(x) = 2/x³

The expression f"(1/x) is quite wrong. I think you should say:
Given the function f(x) = 1/x, what is f"(x)? That is more appropriate.^_^
^_^
^_^

Answer 5

Use the following rule to differentiate a function:
f(x) = x^n => f'(x) = nx^(n-1)

f(x) = 1/x = x^-1

=> f'(x) = -1x^-2
=> f''(x) = 2x^-3

or f"(x) = 2/(x^3)

Answer 6

F'(x)=d/dx(1/x)=-1/(x^2)

f''(x)=2/x^3 which is your answer

Answer 7

f(x) = 1/x = x^(-1)
f'(x) = (-1)*x^(-1-1) = - x^(-2)
f''(x) = - (-2)* x^(-2-1) = 2(x^-3) = 2/(x^3)

Using f(x^n) = nf(x^(n-1))

Answer 8

f** =hu-56 x gh/45 = hy=67

Answer 9

f (x) = 1/x
f ' (x) = -1/x^2
f " (x) = 2/x^3

Answer 10

f''(1/x) = d/dx(d/dx(1/x)) = d/dx(d/dx(x^-1)
= d/dx(-(x^-2))
= -(-2)x^-3 or 2 x^-3 or 2/x^3