Find f '(x) and f '(c).
f(x) = (x^3 + 4x)(2x^4 + 2x − 2), c = 0
f '(x) = 2(7x^6+20x^4+4x^3-3x^2+8x-4)
what is the derivative of c?
I put:
C
0
1
they were all wrong.
the derivative of c= ?
2015-07-31 3:29 am
回答 (2)
2015-07-31 3:40 am
f(x) = (x^3 + 4x)(2x^4 + 2x − 2)
f(x) = 2x^7 + 8x^5 + 2x^4 - 2x^3 + 8x^2 - 8x
f '(x) = 14x^6 + 40x^4 + 8x^3 - 6x^2 + 16x - 8
At c = 0, f '(0) = 14(0^6) + 40(0^4) + 8(0^3) - 6(0^2) + 16(0) - 8 = -8
f(x) = 2x^7 + 8x^5 + 2x^4 - 2x^3 + 8x^2 - 8x
f '(x) = 14x^6 + 40x^4 + 8x^3 - 6x^2 + 16x - 8
At c = 0, f '(0) = 14(0^6) + 40(0^4) + 8(0^3) - 6(0^2) + 16(0) - 8 = -8
2015-07-31 3:39 am
f(x) = (x^3 + 4x)(2x^4 + 2x − 2)
So, the derivative of this is:
(3x^2 + 4)(2x^4 + 2x − 2) + (x^3 + 4x)(8x^3 + 2)
I think you confused the question. It's not asking for the derivative of c, but rather f'(c). Plug c into the derivative:
(0+4)(0+0-2) + (0), should be -8.
So, the derivative of this is:
(3x^2 + 4)(2x^4 + 2x − 2) + (x^3 + 4x)(8x^3 + 2)
I think you confused the question. It's not asking for the derivative of c, but rather f'(c). Plug c into the derivative:
(0+4)(0+0-2) + (0), should be -8.
收錄日期: 2021-04-21 12:02:29
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