How do you differentiate #f(x)=8x(2x+1)^2 + 3(2x+1)^3# using the product rule?

Answer 1

Charles Anctil

#f'(x)=168x^2+136x+26#

Find the derivative of each part.

#d/dx[8x(2x+1)^2]=(2x+1)^2d/dx[8x]+8xd/dx[(2x+1)^2]#

#=8(2x+1)^2+16x(2x+1)d/dx[2x+1]#

#=8(2x+1)^2+32x(2x+1)#

#=96x^2+64x+8#

#d/dx[3(2x+1)^3]=9(2x+1)^2d/dx[2x+1]#

#=18(2x+1)^2#

#=72x^2+72x+18#

Add the two derivatives we've found:

#f'(x)=96x^2+64x+8+72x^2+72x+18#

#=168x^2+136x+26#

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Answer 2

Luke Alvarez

To differentiate the function ( f(x) = 8x(2x+1)^2 + 3(2x+1)^3 ) using the product rule, you would first identify two functions being multiplied together: ( u(x) = 8x ) and ( v(x) = (2x+1)^2 + 3(2x+1)^3 ). Then, you would apply the product rule formula:

[ (uv)' = u'v + uv' ]

Taking the derivatives:

[ u'(x) = 8 ] [ v'(x) = 2(2x+1) \cdot 2 + 3 \cdot 3(2x+1)^2 ]

Then, apply the product rule:

[ f'(x) = 8(2x+1)^2 + 8x \cdot 2(2x+1) + 3(2x+1)^3 + 3 \cdot 3(2x+1)^2 ]

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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.