And explain, don't just tell me the answer because I know it I just don't know how to get it
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answers:
Jessica say: henot sure what to say about this
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Linda K say: by power rule: 3(3+2x - x^2)^2 then * the derivative of inside of ( ) by chain rule
3(3 +2x -x^2)^2 (2 -2x)
3(3+2x -x^2)^2 *2(1-x)
6(1-x)(3+2x - x^2)^2 you can factor the quadratic
6(1-x)((1 -x)(3-x))^2
6(1-x)^3(3-x)^2
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Deepak Suwalka say: y = ( 3 + 2x - x² )³
Using power and chain rule we get:
dy/dx = 3(3 + 2x - x²)² × d/dx (3 + 2x - x²)
= 3(2 - 2x)( 3 + 2x - x² )²
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Sqdancefan say: Use the power and chain rules.
.. d(u^3)/dx = 3u^2*(du/dx)
For u = 3 +2x -x^2, du/dx = 2 -2x and the derivative is
.. 3*(3 +2x -x^2)^2*(2 -2x)
.. = 6*(1 -x)*(3 +2x -x^2)^2
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Steve A say: Let (3+2x-x^2) = u
Derivative u^3 = 3u^2 du
Thus, 3(3+2x-x^2)^2(2 - 2x)
(6 - 6x)(3 + 2x - x^2)^2
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? say: y=(3+2x-x²)³
power rule + chain rule
y'=3(3+2x-x²)²•(3+2x-x²)'
= 3(3+2x-x²)²(2-2x)
= 3(x²-2x-3)²•-2(x-1)
= -6(x-1)((x-3)(x+1))²
= -6(x-1)(x+1)²(x-3)²
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az_lender say: y = (3+2x-x^2)^3, call this u^3.
dy/dx = (dy/du)(du/dx)
= (3u^2)(2 - 4x)
= 3(3+2x-x^2)^2 * (2-4x).
It is not necessary to carry out the multiplication.
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J say: (3+2x-x^2)^3 = -x^6 + 6 x^5 - 3 x^4 - 28 x^3 + 9 x^2 + 54 x + 27
now just differentiate term by term
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Jim Moor say: Here:
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hayharbr say: If it was just x^3, you'd get 3x^2. But you can't just say this derivative is
3(3 + 2x - x^2) ^ 2
When there is more than just x, you have to do the derivative of what's inside the parentheses and multiply by the 3(3 + 2x - x^2) ^ 2
so the final answer is 3(2 - 2x)(3 + 2x - x^2)^2
It's the chain rule:
let u = 3 + 2x - x^2
and y = u^3
You want dy/dx but can only find dy/du and du/dx
But that's ok because dy/dx = dy/du • du/dx
so since du/dx = 2 - 2x, and dy/du = 3u^2, dy/dx = (2 - 2x)(3u^2) and subbing back for u, you get
(2 - 2x)(3(3 + 2x - x^2)^2)
or 3(2 - 2x)(3 + 2x - x^2)^2
That's how I learned it anyway.
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Kara say: heNot sure what to say about this
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