# How do you find the derivative of #y=3(x^2+1)(2x^2-1)(2x+3)#?

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After the formula

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The product rule will be applied in this case:

The expression becomes: after it is simplified.

I hope that was clear.

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To find the derivative of ( y = 3(x^2+1)(2x^2-1)(2x+3) ), you would apply the product rule, which states that the derivative of a product of functions is the sum of the derivative of each function times the other functions. Using the product rule, the derivative of the given function is:

[ y' = 3[(x^2+1)(4x(2x^2-1)(2x+3)) + (x^2+1)(2x^2-1)(4x(2x+3))] ]

Simplify the expression:

[ y' = 3[(4x^3+4x)(2x^2-1)(2x+3) + (2x^2-1)(4x)(2x+3)] ]

[ y' = 3[(8x^5+12x^4-4x^3+8x^3-12x^2+4x)+(8x^3-4x^2+12x^2-6x)] ]

[ y' = 3[8x^5+12x^4+4x-12x^2+8x^3-12x^2+8x^3-4x^2+12x^2-6x] ]

[ y' = 3[8x^5+24x^3-16x^2-6x] ]

[ y' = 24x^3+72x^3-48x^2-18x ]

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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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