Question

How do you differentiate the given function? F(y) = (`1/y^2` - `3/y^4`)(y + 5y^3)

`3/y^4`

Answer 1

`F'(y) = 14/y^2 + 9/y^4 + 5`

Explanation:

STEP 1: Simplify the equation by rewriting with negative exponents. For example, `1/y^2` can be written as `y^-2`.

Your simplified equation should look at follows:
`F(y) = (y^-2 - 3y^-4)(y + 5y^3)`

STEP 2: In order to take the derivative for this problem, we will need to use the Product Rule.
(in case you forgot: `d/dx f(x)*g(x) = f'(x)g(x) + f(x)g'(x)`)
You can set f(y) equal to (`y^-2` - `3y^-4`) and g(y) equal to (y + `5y^3`)

STEP 3: You can now begin to differentiate the expression. Before simplification, `F'(y) = (-2y^-3 + 12y^-5)(y + 5y^3) + (y^-2 - 3y^-4)(1 + 15y^2)`.

STEP 4: You can now expand and simplify the equation!

`F'(y) = 58y^-2 - 10 + 12y^-4 + y^-2 + 15 - 3y^-4 - 45y^-2 + 5`

`F'(y) = 59y^-2 + 9y^-4 + 5`

`F'(y) = 14/y^2 + 9/y^4 + 5`

Answer 2

`f’(y)=5+(14/y^2)+(9/y^4)`

Explanation:

`f(y)=(1/y)+5y-(3/y^3)-(15/y)`
`f(y)=5y-(14/y)-(3/y^3)`

`f’(y)=5+(14/y^2)+(9/y^4)`