derivitive y= (x^3)(2x-5)^4?

Question

using chain rule and sum and difference rule combination i get
8x^3(2x-5)^3+3x^2(2x-5)^4
the answer is in the form of
x^2(2x-5)^3+(14x-5)
where am i going wrong?

Answer 1

y= (x^3)(2x-5)^4

dy/ dx = (x^3 )( 4)( 2x - 5 )^3 ( 2 ) + (2 x - 5 )^4 ( 3x^2 )
= 8 x^3 ( 2x - 5)^3 + 3x^2 ( 2x - 5)^4

you can take out heighest common factor

= x^2(2x - 5 )^3 [ 8x + 3( 2x - 5 ) ]
= x^2(2x - 5 )^3 [ 8x + 6x - 15 ]
= x^2(2x - 5 )^3( 14 x - 15 )

Answer 2

Chain rule and product rule to the rescue!

dy/dx = (3x^2)(2x-5)^4 + (x^3)(4(2x-5)^3)(2)
= x^2 * (2x-5)^3 * (3(2x-5) + 8x) [after factoring]
= x^2 * (2x-5)^3 * (6x -15 + 8x)
= x^2 * (2x-5)^3 * (14x -15)

Answer 3

f `(x) = 3x².(2x - 5)^4 + 4.(2x - 5)³.2.(x³)
f `(x) = 3x².(2x - 5)^4 + 8x³.(2x - 5)³
f `(x) = x².(2x - 5)³.[ 3.(2x - 5) + 8x ]
f `(x) = x².(2x - 5)³.[14x - 15 ]

Answer 4

You copied the answer wrong.
it's x^2[(2x-5)^3(14x-15)]

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