How can i use the product rule to find the derivative to the following function?

Question

f(x)=x^2/5(x-1)

the product rule equation:
if y=uv, then y'=u'v+uv'

Answer 1

if by this you mean

f(x) = (x^(2/5)) * 5(x - 1), then

f'(x) = (2/5)x^((2/5) - 1)*(5(x - 1)) + (x^(2/5))(5(x - 1)')
f'(x) = (10(x - 1)/5)x^((2/5) - (5/5)) + 5x^(2/5)
f'(x) = (2(x - 1))x^(-3/5) + 5x^(2/5)
f'(x) = ((2x - 2)/(x^(3/5)) + 5x^(2/5)
f'(x) = ((2x - 2) + 5(x^(2/5) * x^(3/5)))/(x^(3/5))
f'(x) = (2x - 2 + 5x^((2/5) + (3/5)))/(x^(3/5))
f'(x) = (2x - 2 + 5x^(5/5))/(x^(3/5))
f'(x) = (2x - 2 + 5x)/(x^(3/5))
f'(x) = (7x - 2)/(x^(3/5))

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But if you mean

f(x) = (x^2)/(5(x - 1))

f'(x) = (x^2)(5(x - 1))^(-1)
f'(x) = (2x^(2 - 1))((1/5)(x - 1)^(-1)) + (x^2)((-1/5)(x - 1)^(-1 - 1))
f'(x) = ((2x)/(5(x - 1)) + ((x^2)/(5(x - 1)^2))

ANS : f'(x) = ((2x)/(5(x - 1)) - ((x^2)/(5(x - 1)^2)

or you can do it like this

f'(x) = (2x(x - 1) - x^2)/(5(x - 1)^2)
f'(x) = (2x^2 - 2x - x^2)/(5(x - 1)^2)
f'(x) = (x^2 - 2x)/(5(x - 1)^2)

or

f'(x) = (x^2 - 2x)/(5(x - 1)(x - 1))
f'(x) = (x^2 - 2x)/(5(x^2 - 2x + 1))
f'(x) = (x^2 - 2x)/(5x^2 - 10x + 5)

ANS : (x^2 - 2x)/(5x^2 - 10x + 5)

Answer 2

f(x) = x^2/5(x-1)

Well, if you have to use the product rule and not the quotient rule, then make f(x) look like:

f(x) = (1/5)x^2(x-1)^(-1)
f'(x) = (2/5)x(x-1)^(-1) + (1/5)x^2(-1(x-1)^(-2)(1))
f'(x) = (1/5) * [(2x)/(x-1) - (x^2)/(x-1)^2]

If you do want to use the quotient rule though, this is the formula:
The quotient rule is sometimes hard to remember, but my calculus teacher thought up a really easy way to remember.
(LoDhi - HiDlo)/Lo^2 (low-D-high minus high-D-low over low squared)
The actual formula is:
f(x) = u/v
f'(x) = (vu' - uv')/ v^2

Answer 3

d/dx=(x^2)(5x-5)^-1, then chain rule/power rule 2x(5x-5)^-1+(-1)(5x-5)^-2(5)(x^2). do ur own dam hw