Pls help,...i know how to find firt and second derivative but dont know how to apply 1 and 0 in this question..
f(x)=x^4+x^3-square root(x)
Find :
f'(1)
f"(0)
Maths~derivative question?
2016-04-15 4:13 pm
回答 (2)
2016-04-15 4:54 pm
✔ 最佳答案
f(x) = x⁴ + x³ - √xf'(x) = 4x³ + 3x² - 1/(2√x)
f"(x) = 12x² + 6x + 1/(4x√x)
f'(1) = 4(1)³ + 3(1)² - 1/[2√(1)] = 13/2
If x = 0, 1/(4x√x) does not exist.
f"(0) does not exist
2016-04-17 10:56 am
f(x) = x⁴ + x³ - √x
f'(x) = 4x³ + 3x² - 1/(2√x)
f"(x) = 12x² + 6x + 1/(4x√x)
f'(1) = 4(1)³ + 3(1)² - 1/[2√(1)] = 13/2
If x = 0, 1/(4x√x) does not exist.
f"(0) does not exist
f'(x) = 4x³ + 3x² - 1/(2√x)
f"(x) = 12x² + 6x + 1/(4x√x)
f'(1) = 4(1)³ + 3(1)² - 1/[2√(1)] = 13/2
If x = 0, 1/(4x√x) does not exist.
f"(0) does not exist
收錄日期: 2021-04-18 14:45:00
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160415081335AApJ3Ky