Whose derivatives are 1 / x ^ 2, 1 / x ^ 3, 2 ^ x, E / x

Whose derivatives are 1 / x ^ 2, 1 / x ^ 3, 2 ^ x, E / x

The meaning of this problem is to integrate the three functions of 1 / x ^ 2, 1 / x ^ 3, 2 ^ X and E / X. you can check the integral formula. As for the first one, you can regard 1 / x ^ 2 as x ^ (- 2) + C, and the answer is - 1 / x + C. similarly, the second one is x ^ (- 3), and the answer is - 1 / X (- 2). The third one, e is a constant, and you can first put it forward, that is to integrate 1 / X

Find the derivative of the following function. Y = x ^ 3 + log2x, if possible,
Find the derivative of the following function. Y = x ^ 3 + log2x, if possible,

The solution is y = x ^ 3 + log2x
Derivation
y‘=（x^3+log2x）’
=（x^3）’+（log2x）’
=3x^2+1/（xln2）

The waist length of an isosceles right triangle 3. How many square decimeters is the area of this triangle?

3.5 × 3.5 × half = 6.125 square decimeter

Given that points a and B are two points on hyperbola y = - 8 / X respectively, three points a, O and B are collinear, passing through point B, make a vertical line of X axis, and the perpendicular foot is C, then calculate the area of triangle AOC

Let a (x, - 8 / x) then B (- x, 8 / x) C (x, 0)
S three AOC = 0.5 * | - 8 / X | * | - x | = 4

How to deduce the derivative of F (x) = 5 ^ (LN (Tan (x))

(a^x)'=a^xlna
[5^(lntanx)]'
=5^(lntanx)ln5(lntanx)'
=[5^(lntanx)ln5/tanx]sec²x
=5^(lntanx)ln5/(sinxcosx)

Fill in the word group () Xin () Ku
Fill in a pair of synonyms

Bear hardships
go through untold hardships

In the rectangular coordinate system, a (- 3,4), B (- 1, - 2), O is the origin. (1) calculate the area of △ AOB; (2) translate the triangle up three unit lengths to get △ a ′ o ′ B ′, and then make the symmetry figure △ a ″ o ″ B ′ of △ a ″ o ″ B ″ about the Y axis. Try to write out the coordinates of each vertex of △ a ′ o ′ B ′ and △ a ″ o ″ B ″

(1) S △ AOB = 6 × 3-12 × 3 × 4-12 × 2 × 6-12 × 1 × 2 = 5; (2) ∵ a (- 3,4), B (- 1, - 2), O (0,0), ∵ a ′ (− 3,4 + 3); B ′ (− 1,3 − 2); O ′ (0,3); ∵ a ″ o ″ B ″, ∵ a ″ (3,4 + 3); B ″ (1,3 − 2); O ″ (0,3)

Let ABC be the length of three sides of triangle ABC, and let the quadratic function y = (a + b) x & # 178; - 2cx - (a-b), when x = - # 189; there is the minimum value B / 2, we prove that ABC is the length of three sides of triangle ABC

Let the quadratic function y = (a + b) x2 + 2cx - (a-b), where is the length of the three sides of △ ABC, and B ≥ a, B ≥ C, when x = - 1 / 2, the minimum value of this function is - A / 2, then the size relationship of a, B, C is - 2C / 2 (a + b) = - 1 / 2, that is, when C = (a + b) / 2, there is 4 (a + b) (B-A) - 4c24 (a + b) = - A2, we get 2b2-a2 -

Given a > 0, b > 0, prove that B2 / A + A2 / B is greater than or equal to a + B
Take care of it····

b2/a+a2/b=（a^3+b^3)/ab=(a+b)(a^2-ab+b^2)/ab
≥(a+b)(2ab-ab)/ab=(a+b)*ab/ab=(a+b)

As shown in the figure, in ladder ABCD, ∠ ABC + ∠ DCB = 90 ° and BC = 2ad, take AB, CD and ad as sides to make square outside the ladder, and their areas are S1, S2 and S3 respectively, then the relationship between them is as follows______ ．

Let's make de ∥ AB through point D, the ∥ quadrilateral abed is parallelogram, ∥ ad = be, ab = De, ∥ B = ∥ Dec, ∥ ABC + ∥ DCB = 90 degree, ∥ Dec + ∥ DCE = 90 degree. ∥ BC = 2ad, ∥ ad = EC, ∥ EC2 = de2 + DC2, ∥ ad2 = AB2 + DC2, ∥ S3 = S1 + S2