How do you differentiate #f(x)=(x+1)(x+2)(x+3)# using the product rule?

Answer 1

You consider two of the factors as one, doing somewhat of a chain rule here.

In general terms, we can state that a product rule for three terms can be depicted as follows:

#(abc)'=(ab)'c+(ab)c'=(a'b+ab')c+(ab)c'=color(green)(a'bc+ab'c+abc')#
Let's do it for your #f(x)=(x+1)(x+2)(x+3)#:
#(df(x))/(dx)=(1)(x+2)(x+3)+(x+1)(1)(x+3)+(x+1)(x+2)(1)#
#(df(x))/(dx)=color(green)(3x^2+12x+11)#
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Answer 2

To differentiate ( f(x) = (x + 1)(x + 2)(x + 3) ) using the product rule, follow these steps:

  1. Identify the functions ( u ) and ( v ). ( u = (x + 1)(x + 2) ) ( v = x + 3 )

  2. Differentiate ( u ) and ( v ) separately. ( u' = (2x + 3) ) ( v' = 1 )

  3. Apply the product rule: ( (uv)' = u'v + uv' ). ( f'(x) = (2x + 3)(x + 3) + (x + 1)(x + 2)(1) )

  4. Simplify the expression. ( f'(x) = (2x^2 + 9x + 9) + (x^2 + 3x + 2) )

  5. Combine like terms. ( f'(x) = 3x^2 + 12x + 11 )

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Answer from HIX Tutor

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.

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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