Differentiate f (x) = sin^-1(x^4) Find the value of the derivative at x=1/3?
回答 (4)
2016-05-11 5:38 am
✔ 最佳答案
f(x) = sin⁻¹(x⁴)f '(x) = d sin⁻¹(x⁴) / d x
= [d sin⁻¹(x⁴) / d x⁴] * (d x⁴ / d x)
= {1 / √[1 - (x⁴)²]} * (4x³)
= 4x³ / √(1 - x⁸)
The derivative at x = 1/3, f' (1/3)
= 4(1/3)³ / √[1 - (1/3)⁸]
= (4/27) / √(6560/6561)
= (4/27) / [4(√410)/81]
= 3/√410
= 3(√410)/410
https://c2.staticflickr.com/8/7504/26912017436_bcda38d530_o.png
2016-05-11 5:17 am
f(x) = sinˉ¹(x^4)
f'(x) = 1/√[1 - (x^4)²] * 4x³
f'(x) = 4x³/ √(1 - x^8)
f'(1/3) = 4(1/3)³ / √[1 - (1/3)^8)
f'(1/3) = (4/27) / √[1 - (1/6561)]
f'(1/3) = (4/27) / (√6560/6561)
f'(1/3) = 4 / 27(√6560/6561)
f'(1/3) = 0.148
f'(x) = 1/√[1 - (x^4)²] * 4x³
f'(x) = 4x³/ √(1 - x^8)
f'(1/3) = 4(1/3)³ / √[1 - (1/3)^8)
f'(1/3) = (4/27) / √[1 - (1/6561)]
f'(1/3) = (4/27) / (√6560/6561)
f'(1/3) = 4 / 27(√6560/6561)
f'(1/3) = 0.148
2016-05-11 5:12 am
u = x^4 .... du = 4x^3
f ' = du / sqrt(1 - u^2)
f '(x) = 4x^3 / sqrt(1 - x^8)
f '(1/3) = (4/27) / sqrt(1 - 1/3^8) <<< simplify as needed or use a calculator
f ' = du / sqrt(1 - u^2)
f '(x) = 4x^3 / sqrt(1 - x^8)
f '(1/3) = (4/27) / sqrt(1 - 1/3^8) <<< simplify as needed or use a calculator
2016-05-11 5:09 am
The derivative of sin^-1 (x) is 1 / sqrt(1 - x^2)
so...
[ 1 / sqrt(1 - x^2) ] * chain rule
[ 1 / sqrt(1 - x^2) ] * 4x^3
4x^3 / sqrt(1 - x^2)
I'm sure you can plug in 1/3 and get your answer just fine
I got .1571 ...
so...
[ 1 / sqrt(1 - x^2) ] * chain rule
[ 1 / sqrt(1 - x^2) ] * 4x^3
4x^3 / sqrt(1 - x^2)
I'm sure you can plug in 1/3 and get your answer just fine
I got .1571 ...
收錄日期: 2021-04-18 14:49:57
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