Finding Derivatives?

Question

What is the derivative of x? is it 0 or 1? im very confused

problem: Find the second derivative

f(x)= x/x-1 (x over x minus 1)

Answer 1

d/dx ( x ) = 1

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

Use the quotient rule:

dy/dx = [ d/dx ( x ) * (x - 1) - ( x ) * d/dx ( x - 1 ) ] / ( x - 1 )²

dy/dx = [ ( 1 ) ( x - 1 ) - ( x ) ( 1 ) ] / ( x - 1 )²

dy/dx = [ ( x - 1 ) - ( x ) ] / ( x - 1 )²

dy/dx = ( -1 ) / ( x - 1 )²

d²y/dx² = [ d/dx ( -1 ) ( x - 1 )² - ( -1 ) d/dx ( x - 1 )² ] / ( x - 1 )^4

d²y/dx² = [ ( 0 ) ( x - 1 )² - (-1) ( 2 (x - 1) (1) ] / ( x - 1 )^4

d²y/dx² = [ - ( -1 ) ( 2x - 2 ) ] / ( x - 1 )^4

d²y/dx² = [ ( 2x - 2 ) ] / ( x - 1 )^4

d²y/dx² = [ 2 ( x - 1 ) ] / ( x - 1 )^4

d²y/dx² = ( 2 ) / ( x - 1 )³

Answer 2

1st derivative of x is always 1. 2nd derivative of x would be 0, since 1 (the 1st derivative is a constant, not a variable)

for the second problem, use the quotient rule where if h = f/g, then the derivative of h, h' = [gf' - fg'] / g²

So, f'(x) ---> first derivative = [(x-1)(1) - (x)(1)] / (x-1)^2

                   so, f'(x) = -1/(x-1)^2

Now find the second derivative using the chain rule.

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

f"(x) = -(-2) X (x-1)

        = 2(x-1)

        = 2x - 2

Hope this helps!

Answer 3

derivative of x is just 1, remember that derivative of any constant value, meaning any number like 1,2,75,76534 is zero, but the derivative of a variable such as x is one. This does not hold for higher powers of x such as x^2.

f(x) = x/x-1

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

so f'(x) = -1/(x-1)^2 = -(x-1)^-2

so now differentiate again using chain rule

f''(x) = 2(x-1)(1) = 2x-2

hope this helps

Answer 4

f(x)= x/x-1

x/(x-1) = 1 + 1/(x-1)

f'(x) = 0 + -( x -1)^-2

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

f"(x) = -( x -1)^-2

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

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