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How do you differentiate #g(x) =(x+2)^2 (2x)# using the product rule?

Answer 1

Scarlett Allison

#dy/dx=6x^2+16x+8#

Given -

#y=(x+2)^2(2x)#

#dy/dx=[(x+2)^2(2)]+[2x .2(x+2) .1]#

#dy/dx=2(x+2)^2+4x(x+2)#

#dy/dx=2(x+2)^2+4x^2+8x#

#dy/dx=2(x^2+4x+4)+4x^2+8x#

#dy/dx=2x^2+8x+8+4x^2+8x#

#dy/dx=6x^2+16x+8#

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Answer 2

Paisley Johnson

To differentiate ( g(x) = (x + 2)^2 \cdot (2x) ) using the product rule, follow these steps:

Identify the functions ( u(x) ) and ( v(x) ).
- Let ( u(x) = (x + 2)^2 ) and ( v(x) = 2x ).
Find the derivatives ( u'(x) ) and ( v'(x) ).
- ( u'(x) ) can be found using the chain rule, which yields ( u'(x) = 2(x + 2) \cdot 1 = 2(x + 2) ).
- ( v'(x) = 2 ), as the derivative of ( 2x ) with respect to ( x ) is simply ( 2 ).
Apply the product rule.
- ( g'(x) = u'(x)v(x) + u(x)v'(x) ).
- Substituting the values: ( g'(x) = (2(x + 2))(2x) + ((x + 2)^2)(2) ).
- Simplify: ( g'(x) = 4x(x + 2) + 2(x + 2)^2 ).
- Expand and collect like terms if necessary.

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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.