High number dy / DX 1. The velocity of the particle moving along the x-axis is DX / dt = f (x), and the acceleration a of the particle is calculated 2. DX / dy = 1 / y ', find the second derivative d2x / dy2 Why is the second derivative not the derivative of the first derivative?

one
a=d2x/dt2=d(dx/dt)/dt
=d[f(x)]/dt
={d[f(x)]/dx}*(dx/dt)
=f'(x)*f(x)
two
d2x/dy2=d(dx/dy)/dy
=D (1 / y ') / dy [note here that y' is a function of X, so take the derivative of X first]
=[d(1/y')/dx]*(dx/dy)
={-y''/[(y')^2]}*(1/y')
=-y''/[(y')^3]

Find the 1 / x power of limit Lim [(1 + x) divided by the 1 / x power of e], when x tends to 0

Let a = (1 + x) ^ (1 / x ^ 2) / e ^ (1 / x)
Then Lim ln a = Lim ln (1 + x) / x ^ 2 - 1 / X
= lim [ ln(1+x) -x ] /x^2
=- 1 / 2 (lobida's law)
So Lim a = e ^ (- 1 / 2)

Find LIM (x approaches 1) (root sign of X to the power of m-1) / (x-1)

The m-th power of the root sign of X refers to the 1 / m power of X
When X - > 1, x ^ (1 / M) - 1
=[1 + (x-1)] ^ (1 / M) - 1 is equivalent to (x-1) / m
LIM (x approaches 1) (root sign of X to the power of m-1) / (x-1)
=LIM (x approaches 1) (x-1) / M (x-1) = 1 / M
Understand the equivalent infinitesimal carefully. It is very important for limit calculation

When x approaches 0, the limit of the 1 / x power of LIM (1-x)? Want process

Solution:
Original formula = LIM (x → 0) (1-x) ^ (1 / x)
=lim(x→0)(1-x)^(1/x)=(1+(-x))^(1/-x) × （-1）
=lim(x→0)e^(-1)
=1/e

The limit x plus one to the fifth power, the root sign to the seventh power minus one, divided by X X tends to 0

X tends to 0, and there is formula (1 + x) to the power of M = 1 + MX
So when x tends to 0 ((1 + x) to the 5th / 7th power - 1) / x = 5 / 7
The final limit is 5 / 7

Function limit calculation if LIM (x →∞) [5x radical (AX ^ 2 + BX + 1)] = 2. Find a, B

∵lim(x→∞)［5x－√（ax^2＋bx＋1）］
＝lim(x→∞)｛［25x^2－（ax^2＋bx＋1）］/［5x＋√（ax^2＋bx＋1）］｝＝2,
∴a＝25.
Otherwise, the numerator is higher-order infinity relative to the denominator,
‡ LIM (x →∞) {[25X ^ 2 - (AX ^ 2 + BX + 1)] / [5x + √ (AX ^ 2 + BX + 1)] = ∞, not 2
From a = 25, we get:
lim(x→∞)［5x－√（ax^2＋bx＋1）］
＝lim(x→∞)｛－（bx＋1）/［5x＋√（25x^2＋bx＋1）］｝
＝－lim(x→∞)｛（b＋1/x）/［5＋√（25＋b/x＋1/x^2）］｝
＝－（b＋0）/［5＋√（25＋0＋0）］
＝－b/10
＝2,
∴b＝－20.
The values of a and B satisfying the conditions are 25 and - 20 respectively

High number dy / DX 1. The velocity of the particle moving along the x-axis is DX / dt = f (x), and the acceleration a of the particle is calculated 2. DX / dy = 1 / y ', find the second derivative d2x / dy2 Why is the second derivative not the derivative of the first derivative?