# How do you differentiate #f(x)= (2x^2-5)(4x^2+5) # using the product rule?

If you are studying maths, then you should learn the Product Rule for Differentiation, and practice how to use it:

I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".

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To differentiate ( f(x) = (2x^2 - 5)(4x^2 + 5) ) using the product rule, we use the formula ( (uv)' = u'v + uv' ). Letting ( u = 2x^2 - 5 ) and ( v = 4x^2 + 5 ), we find ( u' = 4x ) and ( v' = 8x ). Applying the product rule, we have:

[ f'(x) = (2x^2 - 5)'(4x^2 + 5) + (2x^2 - 5)(4x^2 + 5)' ] [ = (4x)(4x^2 + 5) + (2x^2 - 5)(8x) ] [ = 16x^3 + 20x + 16x^3 - 40x ] [ = 32x^3 - 20x ]

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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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