DX / dy = 1 / y ', find D square X / dy square

DX / dy = 1 / y ', find D square X / dy square

Here, we take x = g (y), X is the dependent variable and Y is the independent variable to find the second derivative of function x with respect to the independent variable y
It is known that the condition DX / dy = 1 / y 'is the derivative relationship between the function x = g (y) and its inverse function y = f (x). The topic means to find the second derivative of function x with respect to independent variable y from this condition
The key to solve this problem is to pay attention to which variable to derive; The derivation method of composite function is used
The specific answers are as follows:
d^2x/dy^2
=D [DX / dy] / Dy (find another derivative for the first derivative)
=D [1 / y '] / Dy (substitution condition)
={d [1 / y '] / DX} * [DX / dy] (because y' in 1 / y 'is the derivative of function y = f (x) and is a function of X, of course, 1 / y' is also a function of X. now if the function of X needs to derive from y, it needs the derivation method of composite function. First derive from X and then y for 1 / y ')
={[- 1 / y '^ 2] * y' '} * [DX / dy] (here {[- 1 / y' ^ 2] * y ''} is obtained again by the derivation method of composite function: first derive y 'from [1 / y'], and then derive y 'from x)
={[- 1 / y '^ 2] * y' '} * [1 / y'] (substitution condition)
=-y''/（y')^3.

Advanced mathematics problems. Solving differential equations (1 + x) ²) dy/dx=2xy-3x ² y ² General solution of

This is a simple question type in the textbook. According to the formula step by step, it is easy to get:
The process is not written. It's all in the book,

Let y = ln [ln (LNX)] solve the problem of dy / DX greater one higher number

This is a composite function, and the derivation only needs to use the derivation rule of composite function, i.e. chain rule:
dy/dx = {ln[ln(lnx)]}′
= 1/[ln(lnx)]·[ln(lnx)]′
= 1/[ln(lnx)]·1/(lnx)·(lnx)′
= 1/{x·(lnx)·[ln(lnx)]}
If you can't understand the above process, you can do the same
Decompose the function y = ln [ln (LNX)] into y = ln u, u = LNv, v = LNX
dy/dx = (dy/du)·(du/dv)·(dv/dx)
= (1/u)·(1/v)·(1/x)
= 1/{x·(lnx)·[ln(lnx)]}

The derivative of F (x) = (x ^ 2 + 2) e ^ x is equal to?

f'(x)=(x^2+2)'e^x+(x^2+2)(e^x)'
=2xe^x+(x^2+2)e^x
=(x^2+2x+2)e^x

How to find the definite integral of the X-Power of E from 0 to 1, that is, what derivative is equal to the X-Power of E

This definite integral cannot be obtained

Seek advice: find the derivative of y = x ^ (x ^ 2). Using the logarithmic derivative method, it is the x square power of X,

y=x^(x^2)
Take the natural logarithm on both sides at the same time to obtain:
lny=(x^2)lnx
When both sides derive x at the same time:
y '/y=(x^2) 'lnx+(x^2)·(lnx) '
y '/y=2xlnx+x
y '=y(2xlnx+x)
Substitute y = x ^ (x ^ 2) into the above formula to obtain:
y '=x^(x^2)·(2xlnx+x)