Find the derivative of the following function y = 1 / X (2 + 5x) ∧ 10

Find the derivative of the following function y = 1 / X (2 + 5x) ∧ 10

y=1/x*(2+5x)∧10
y'=-1/x^2*(2+5x)^10+1/x*10(2+5x)^9*5
=(2+5x)^9/x^2*[50x-(2+5x)]
=(2+5x)^9*(45x-2)/x^2
Given that function f (x) is an odd function defined on R, f (1) = 0, XF '(x) - f (x) > 0 (x > 0), then the solution set of inequality x ^ 2F (x) > 0 is
The construction function g (x) = f (x) / x, and its derivation should be: the denominator is the square of X, and the molecule is XF '(x) - f (x),
We know that XF '(x) - f (x) > 0 (x > 0), that is to say, G (x) is an increasing function when x > 0, and because g (1) = 0, there is a condition that x is not equal to 0 in the solution set of 00 compared with that of F (x) > 0!
This kind of topic is very common. You should summarize it more,
The derivative of F (x) is, XF '(x) - f (x), x > 0, XF' (x) - f (x) > 0, y = XF (x) is an increasing function on x > 0
F (x) is an odd function defined on R, y = XF (x) is an even function, f (1) = 0, f (- 1) = 0
The solution set of inequality XF (x) > 0 is (negative infinity, - 1) and (1, positive infinity)
Finding the derivative of function y = a ^ x + sin (5x-1)
solution
y'=a^xIna+5cos[5x-1]
formula
a^x'=a^xIna
I don't know how to ask
y‘=(a^x)lna+5cos(5x-1)
Have a good time! Hope to help you, if you do not understand, please ask, I wish learning progress! O(∩_ ∩)O
y′=a^x lna+5cos(5x-1)
Let f (x) be an odd function defined on R, with XF '(x) + F (x) < 0 and f (- 2) = 0 on (- ∞, 0), then the solution set of the inequality XF (x) < 0 is ()
A. {x | - 2 ＜ x ＜ 0 or X ＞ 2} B. {x | - 2 or 0 ＜ x ＜ 2} C. {x | - 2 or X ＞ 2} D. {x | - 2 ＜ x ＜ 0 or 0 ＜ x ＜ 2}
Let g (x) = XF (x), then G '(x) = [XF (x)]' = x'f (x) + XF '(x) = XF' (x) + F (x) ＜ 0, the function g (x) is a decreasing function in the interval (∞, 0), the function f (x) is an odd function defined in R, the function g (x) = XF (x) is an even function in R, and the function g (x) is a decreasing function in the interval (∞, 0)
Finding the cubic power of y = 2x - the square of 3x - 12x + 20 finding monotone interval and extremum
·····
Monotone increasing intervals: (- ∞, - 2) and (1, ∞)
Monotone decreasing interval: (- 2,1)
Maximum: x = - 2, f (x) = 20;
Minimum: x = 1, f (x) = - 7;
If you want to process, please contact me!
Do it with derivatives. Derivation of X
If the odd function f (x) defined on R is an increasing function on (0, + ∞), and f (- 3) = 0, then the solution set of the inequality XF (x) < 0 is ()
A. （-3，0）∪（0，3）B. （-∞，-3）∪（3，+∞）C. （-3，0）∪（3，+∞）D. （-∞，-3）∪（0，3）
It is shown that: F (- 3) = - f (3) = 0, f (3) = 0, and f (x) is an increasing function on (0, + ∞). When 0 ＜ x ＜ 3, f (x) ＜ 0, when x ＞ 3, f (x) ＞ 0, and f (x) is an odd function defined on R, f (- 3) = 0, when x ＜ - 3, f (x) ＜ 0, when - 3 ＜ x ＜ 0, f (x) ＞ 0. The picture is as follows: the solution set of inequality XF (x) ＜ 0 is So we choose a
Extreme value of y = (3x ^ 3-4x) (2x + 1)
dy/dx=(3x^3-4x)'(2x+1)+(3x^3-4x)(2x+1)'
=(9x^2-4)(2x+1)+(3x^3-4x)(2)
=(18x^3-8x+9x^2-4)+(6X^3-8x)
=24x^3+9x^2-16x-4
Or y = (3x ^ 3-4x) (2x + 1) = 6x ^ 4-8x ^ 2 + 3x ^ 3-4x
dy/dx=24x^3-16x+9x^2-4
Given that f (x) is an odd function defined on (- 1,1) and monotonically decreasing in the domain of definition, the inequality f (x-1) + F (2x-1)
f(x-1)+f(2x-1)
F (x) is an odd function defined on (- 1,1), f (2x-1) = - f (1-2x)
f(x-1)+f(2x-1)2/3
There is also domain - 1
① Find the monotone interval and extremum of F (x) = x-3x Λ 2. 2. Given y = 4x Λ 2-2x, find y '
③ Given y = xlnx, find y '(E)
④ Y = arcsin (1-x Λ 2), find dy
(1) F (x) = x-3x ^ 2 = - 3 * (x ^ 2-x / 3) = - 3 (x ^ 2-x / 3 + 1 / 36) + 3 / 36 = - 3 (x-1 / 6) ^ 2 + 1 / 12, so, when x 1 / 6, there is a monotone decreasing; when x = 1 / 6, there is a maximum, which is also the maximum, 1 / 12 (2), y '= 8x-2 (3), y' = LNX + 1, y '(E) = lne + 1 = 2 (4), Dy = [1 / √ (1 - (1-x ^ 2) ^ 2)] * (- 2x *
① Derivation of F (x)
f'(x)=1-6x
Let f '(x) = 0, x = 1 / 6
So (negative infinity, 1 / 6) interval monotone increasing, (1 / 6, positive infinity) interval monotone decreasing, maximum f (1 / 6) = 1 / 12
②y'=8x-2
3. Answer: 2
4 answer: - 2x / radical (1 - (1-x ^ 2) ^ 2) ask: boss process! The direct answer will be criticized
Let f (x) be a monotone increasing function over the domain (0, positive infinity), and for any x, y in the domain, f (XY) = f (x) + F
All have f (XY) = f (x) + F (y), f (2) = 1, find the inequality f (x) + 2
Remember to adopt it first^^
f(3-x)
≥f(x)+2
=f(x)+1+1
=f(x)+f(2)+f(2)
=f(2x)+f(2)
=f(4x)
That is, f (3-x) ≥ f (4x)
Because monotone increasing function
3-x ≥ 4x, i.e. x ≤ 3 / 5
And ∵ 3-x > 0, x > 0
∴0＜x＜3
So 0 < x ≤ 3 / 5