# How do you differentiate #f(x)= (2x^2-5)(x+1) # using the product rule?

Written informally, this is:

(derivative of product)=(first)x(derivative of second)+(second)x(derivative of the first).

The derivative of the 'second' is calculated:

Using the product rule (and the derivatives worked out above) gives:

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To differentiate ( f(x) = (2x^2 - 5)(x + 1) ) using the product rule, follow these steps:

- Identify the two functions being multiplied: ( u(x) = 2x^2 - 5 ) and ( v(x) = x + 1 ).
- Apply the product rule formula: ( f'(x) = u'(x)v(x) + u(x)v'(x) ).
- Differentiate ( u(x) ) and ( v(x) ) separately:
- ( u'(x) = \frac{d}{dx}(2x^2 - 5) = 4x )
- ( v'(x) = \frac{d}{dx}(x + 1) = 1 ).

- Substitute the derivatives and the original functions into the product rule formula:
- ( f'(x) = (4x)(x + 1) + (2x^2 - 5)(1) ).

- Simplify the expression:
- ( f'(x) = 4x^2 + 4x + 2x^2 - 5 ).

- Combine like terms:
- ( f'(x) = 6x^2 + 4x - 5 ).

So, the derivative of ( f(x) ) with respect to ( x ) using the product rule is ( 6x^2 + 4x - 5 ).

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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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