How do you differentiate the equation? $F(y)= (1/y^2 + 3/y^4) (y+5y^3)$

Question

Differentiate:
$F(y)= (1/y^2 + 3/y^4) (y+5y^3)$

Product Rule

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Answer 1 · 2016-10-11T16:35:58

$F'(y) = - 9/y^4 - 16/y^2 + 5$

The first step in finding the derivative of this equation is to rewrite the equation F(y) as:

$F(y) = (y^-2 + 3y^-4)(y + 5y^3)$
By rewriting the exponents it will be easier to use the product rule in the next step.

Now use the Product Rule to find the derivative.
Product Rule: $d/dx (f(x)g(x)) = d/dx f(x)*g(x) + f(x) * d/dx g(x)$

$F'(y) = (-2y^-3-12y^-5)(y+5y^3) + (y^-2+3y^-4)(1+15y^2)$

Expand the equation by FOILing/distributing:
$F'(y) =-2y^-2 - 10y^0-12y^-4-60y^-2+15y^0+y^-2+45y^-2+3y^-4$

Combine like terms:
$F'(y) = -9y^-4 -16y^-2 +5$

Rewrite the equation:
$F'(y) = - 9/y^4 - 16/y^2 + 5$