How do you differentiate #y=(x+5)(2x3)(3x^2+4)#?
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To differentiate the function ( y = (x+5)(2x3)(3x^2+4) ), you can use the product rule of differentiation:
[ \frac{d}{dx}[f(x)g(x)h(x)] = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x) ]

Find the derivatives of each term:
 ( f(x) = x + 5 ) (first factor)
 ( g(x) = 2x  3 ) (second factor)
 ( h(x) = 3x^2 + 4 ) (third factor)

Find the derivatives of each factor:
 ( f'(x) = 1 ) (derivative of ( x + 5 ))
 ( g'(x) = 2 ) (derivative of ( 2x  3 ))
 ( h'(x) = 6x ) (derivative of ( 3x^2 + 4 ))

Apply the product rule: [ y' = (1)(2x3)(3x^2+4) + (x+5)(2)(3x^2+4) + (x+5)(2x3)(6x) ]

Simplify the expression to obtain the final derivative.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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