How do you differentiate #g(y) =(y^-1 + y^-2)(5y^2 -y^3)# using the product rule?

Answer 1

#g'(y)=4-2y#.

#g(y)=(y^-1+y^-2)(5y^2-y^3)=(y^-1+y^-2)y^2(5-y)=(y^-1*y^2+y^-2*y^2)(5-y)=(y+1)(5-y)#
Using Product Rule #: d/dy(U*V)=U*d/dyV+V*d/dyU#
#g'(y)=(y+1)d/dy(5-y)+(5-y)d/dy(y+1)#
#=(y+1){d/dy5-d/dyy}+(5-y){d/dyy+d/dy1}#
#=(y+1)(0-1)+(5-y)(1+0)=-y-1+5-y=4-2y#

Take note that, in order to proceed, we had to use the Product Rule.

#g(y)=(y+1)(5-y)=5y+5-y^2-y=5+4y-y^2#' so that,
#g'(y)=(5)'+(4y)'-(y^2)'=0+4*1-2y=4-2y#, as before!

I hope this helps! Have fun with math!

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Answer 2

To differentiate ( g(y) = (y^{-1} + y^{-2})(5y^2 - y^3) ) using the product rule, follow these steps:

  1. Identify the functions ( u(y) = y^{-1} + y^{-2} ) and ( v(y) = 5y^2 - y^3 ).
  2. Find the derivatives of ( u(y) ) and ( v(y) ) separately.
  3. Apply the product rule, which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.
  4. Substitute the derivatives and original functions into the product rule formula.
  5. Simplify the expression to get the final result.

Let's do it step by step:

  1. ( u(y) = y^{-1} + y^{-2} ) and ( v(y) = 5y^2 - y^3 ).
  2. Find the derivatives:
    • ( u'(y) = -y^{-2} -2y^{-3} )
    • ( v'(y) = 10y - 3y^2 ).
  3. Apply the product rule: [ g'(y) = u'(y) \cdot v(y) + u(y) \cdot v'(y) ].
  4. Substitute the derivatives and original functions: [ g'(y) = (-y^{-2} -2y^{-3})(5y^2 - y^3) + (y^{-1} + y^{-2})(10y - 3y^2) ].
  5. Simplify the expression: [ g'(y) = (-5y^{-2} + y^{-1} + 10y^{-3} - 2y^{-2})(5y^2 - y^3) + (10y^0 - 3y^{-1} + 10y^{-2} - 3y^{-3}) ]. [ g'(y) = -25y^{-2} + 5y^{-1} + 50y^{-3} - 10y^{-2} + 50y^{-3} - 10y^{-2} + 10 - 3y^{-1} + 10y^{-2} - 3y^{-3} ]. [ g'(y) = 10 - 3y^{-1} - 3y^{-3} ].

So, the derivative of ( g(y) ) with respect to ( y ) is ( 10 - 3y^{-1} - 3y^{-3} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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