find the derivative (y^-2+y^-1)(4y^-3-5y^-4)?

Question

step by step

Answer 1

Let's get rid of the negative exponents:
(1/y^2 +1/y)(4/y^3 -5/y^4)
= [(1+y)/y^2] [(4y-5)/y^4]
= [(1+y)(4y-5)]/y^6
= (4y^2-y-5)/y^6
Now use quotient rule
[y^6(8y-1) -6y^5(4y^2-y-5)]/y^12
(8y^7 -y^6 -24y^7 -6y^6 -30y^5)/y^12
(-16y^7-7y^6 -30y^5)/y^12
-16/y^5 - 7/y^6 -30/y^7

Answer 2

Remember: (x^n)' = n x ^(n-1)
Therefore:
y = (y^-2 + y^-1) (4y^-3 - 5y^-4)

Use the Product rule:
(uv)' = uv' +u'v

y' = (y^-2 +y^-1) (-12y^-4 + 20y^-5) + (-2y^-3 - y^-2) (4y^-3 - 5y^-5)

Then multiply and simplify
Ans:

y' = -2y^-8 +25y^-7 -18y^-5 +12y^-6

Make sure to check!!!

Answer 3

f(y) = (y^-2+y^-1)(4y^-3-5y^-4) = 4y^-4 - y^-5 - 5y^-6
f'(y) = -16y^-5 + 5y^6 +30y^-7