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Let y = f [(3x-2) / (3x + 2)] and f '(x) = arctanx ^ 2, then what is the value of dy / DX x = 0 Why do I bring [(3x-2) / (3x + 2)] directly into arctanx ^ 2 Is arctan [(3x-2) / (3x + 2)] ^ 2|x = 0 The results are different?
dy/dx|x=0 =df[(3x-2)/(3x+2)]/dx|x=0 =arctan[(3x-2)/(3x+2)]^2*[(3x-2)/(3x+2)]'|x=0
=3π/4
If f '(x) = SiN x ^ 2, y = f (2x / x-1), find dy / DX
dy/dx = y'
= f'[ 2x / (x-1) ] * [ [ 2x / (x-1) ] '
= sin[ 2x / (x-1) ] ² * - 2 / (x-1) ²
= -2 sin[ 2x / (x-1) ] ² / (x-1) ²
Y = f [(x-1) / (x + 1)], f '(x) = arctanx ^ 2, find dy / DX, dy
Y = f [(x-1) / (x + 1)], f '(x) = arctanx ^ 2, find dy / DX, and derive x on both sides of Dy: dy / DX = f' [(x-1) / (x + 1)] * 2 / (x + 1) ^ 2 = arctan [(x-1) / (x + 1)] ^ 2 * 2 / (x + 1) ^ 2dy = f '[(x-1) / (x + 1)] * 2 / (x + 1) ^ 2 = arctan [(x-1) / (x + 1)] ^ 2 * 2 / (x + 1) ^ 2 * DX
Find the monotone interval and extreme value of function f (x) = 2x ^ 2-lnx
F (x) = 2x ^ 2-lnx, the domain is x > 0f '(x) = 4x-1 / x, Let f' (x) = 0,4x-1 / x = 0 get x = 1 / 2 or - 1 / 2. Because x > 0, the extreme value is x = 1 / 2F "(x) = 4 + 1 / x ^ 2. When x = 1 / 2, F" (1 / 2) = 8 > 0, so x = 1 / 2 is the minimum. F (x) decreases monotonically in the (0,1 / 2) interval and increases monotonically in the [1 / 2, positive infinity) interval
The decreasing interval of the function y = 2x LNX is _
∵ the definition field of y = 2x LNX is (0, + ∞) ∵ y '= 2-1
x
Order 2-1
X < 0, get 0 < x < 1
two
So the answer is: (0, 1)
2)
The monotone decreasing interval of function f (x) = 2x2 LNX is _
From F (x) = 2x2 LNX, we get: F ′ (x) = (2x2 LNX) ′ = 4x-1x = (2x + 1) (2x-1) X. because the definition field of function f (x) = 2x2 LNX is (0, + ∞), from F ′ (x) ＜ 0, we get: (2x + 1) (2x-1) x ＜ 0, that is, (2x + 1) (2x-1) ＜ 0, the solution is: 0 ＜ x ＜ 12. The function f (x)
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