Two problems of high numbers calculus Z = x ^ y is partial to y, and the value at (1 / x, 1) is DZ / dy │ (1 / x, 1) = () Another question is Z = cot (2XY ^ 2), then ә z/ ә y=( )

Two problems of high numbers calculus Z = x ^ y is partial to y, and the value at (1 / x, 1) is DZ / dy │ (1 / x, 1) = () Another question is Z = cot (2XY ^ 2), then ә z/ ә y=( )

lnZ=ylnx
( ә z/ ә y)/Z=lnx
So, ә z/ ә y=x^y*lnx
So, ә z/ ә y│(1/x ,1)=(-lnx)/x
ә z/ ә y=-(csc(2xy^2))^2*4xy

Find the solution of 2 calculus problems (including the process), 1.integrate [(x^2-25)^(1/2)]/(x^4) dx 2.integrate [16x^2-25]^(-3/2) dx

1.∫ [∫√(x ²- 25)] / (x ⁴) DX order x ²= u. Then x = √ u, DX = Du / 2x = (1 / 2) Du / (√ U), substitute into the original formula to obtain: original formula = (1 / 2) ∫ [√ (u-25)] / [(U) ² √u)]du=(1/2)∫[√(1-25/u)]du/u ²= (1/50)∫[√(1-25/u)]d[1-(25/u)]=(1/50)...

Find the definite integral of function 1 / [(1-cosx) (SiNx) ^ 2] on 2 * arctan2 to π / 4 for detailed explanation

Let t = Tan (x / 2), then DX = 2DT / (1 + T) ²), sin(x/2)=t/√(1+t ²), cos(x/2)=1/√(1+t ²) Thus, the original formula = 1 / 4 ∫ (2, Tan (π / 8)) (1 + 2 / T ²+ 1/t^4)dt =1/4(t-2/t-(1/3)/t ³) │(2,tan(π...

A mathematical limit problem According to the definition, it is proved that when x tends to zero, y = (1 + 2x) / X is infinite. What conditions should x meet to make the absolute value of Y > 10 ^ 4?

When x belongs to negative infinity to - 1 / 2, the absolute value of y = (1 + 2x) / x, and then we get a set of inequalities (1 + 2x) / x greater than the fourth power of 10, X less than - 1 / 2. Note that the inequality is the same as × Negative numbers change sign, then calculate no solution, and then discuss that x belongs to - 1 / 2 to 0. At this time, y absolute value = - (1 + 2x) / x, and then the same inequality system still has no solution. Finally, discuss that x belongs to 0 to positive infinity, then y absolute value (1 + 2x) / X is greater than the fourth power of 10, X is greater than 0, 1 / x + 2 is greater than 10000, 1 / X is greater than 9998, so x is greater than 0 and less than 1 / 9998. In conclusion, X is greater than 0 and less than 1 / 9998

Thank you very much! P. S: I want to understand how this type of problem is done. Thank you! If the process is complete and clear, add points! 1、lim = [(x + h)^3-x^3] / h (x - 0) 2、lim = sin2x / sin7x (x - 0) 3、lim = (1 - cos2x) / (xsinx) (x - 0) 4. Lim2 ^ n * sin (x / 2 ^ n) (x is not equal to 0) (n - infinite) 5、lim[(1 + 2)/x]^(x + 3) (n - infinite) 6、lim[(1 + x) / x]^2x (x - infinite)

1、lim = [(x + h)^3-x^3] / h
(x - 0)
（x + h)^3-x^3=[(x+h)-x][(x+h)^2+ x(x+h)+ h^2]
The original formula is equal to Lim [(x + H) ^ 3-x ^ 3] / h
(x - 0)
= lim [(x + h)^2+x(x+h)+h^2]
(x - 0)
=2h^2
2、lim = sin2x / sin7x
(x - 0)
Equivalent Infinitesimal Substitution formula SiNx ~ x (when x tends to 0)
The original formula is equal to
lim sin2x / sin7x
(x - 0)
=2x/7x
=2/7
3、lim (1 - cos2x) / (xsinx)
(x - 0)
Replacement formula of equivalent infinitesimal (1) SiNx ~ x (2) 1-cosx ~ x ^ 2 / 2 (when x tends to 0)
The original formula is equal to
lim (1 - cos2x) / (xsinx)
(x - 0)
=lim [(2x)^2]/2/x^2
=2
4. Lim 2 ^ n * sin (x / 2 ^ n) (x is not equal to 0)
(n - ∞)
=lim sin(x/2^n)
（n-∞）--------- * x
x/2^n
When n - ∞, X / 2 ^ n (here x is regarded as a constant) tends to 0
Equivalent Infinitesimal Substitution formula SiNx ~ x (when x tends to 0)
The original formula is equal to
=x
5、lim[1 + 2/x]^(x + 3)
(x - infinite)
This type corresponds to 1 ∞ type, and the formula LIM (1 + 1 / x) ^ (x) = E
The original formula is equal to=
lim[1 + 2/x]^x * lim[1 + 2/x]^3
(x-∞） (x-∞)
=lim{[1+1/(x/2)]^(x/2)}^2 *1
=e^2
6、lim[(1 + x) / x]^2x
(x - infinite)
This type corresponds to 1 ∞ type, and the formula LIM (1 + 1 / x) ^ (x) = E
(x-∞)
Original formula = Lim [(1 + 1 / x) ^ x] 2
(x - infinite
=e^2

Function problem, do me a favor Is y = 3 a positive proportional function? Explain why

The definition of positive proportional function is y = KX! This is not called a positive proportional function, but a constant value function!