Y = x ^ (x ^ 2) derivation

Y = x ^ (x ^ 2) derivation

Take logarithm on both sides
lny=x^2*ln|x|
Two side derivation
y'/y=2xln|x|+x^2/x
y'=y(2xln|x|+x)=x^(x^2)(2xln|x|+x)

Derivative y ^ 2 = x + 4 If you can, please attach process knowledge points and so on

If the equation f (x, y) = 0 can determine the corresponding relationship between Y and X, then this equation is called an implicit function. This is the shape of X-Y ^ 2 + 4 = 0. For example, there are two methods for this implicit function, one is to divide Y > 0 and Y < 0 into y = ±√ (x + 4) (this is called the explicit derivation of the implicit function), and then find the derivative. But not all the implicit functions

Y = x / (x ^ 2 + 3) derivation

y=x/(x ²+ 3)
y'=1/(x ²+ 3) ²· [(x ²+ 3)-x(2x)]
=1/(x ²+ 3) ²· (x ²+ 3-2x ²)
=(3-x ²)/ (x ²+ 3) ²
Quotient rule: (U / V) '= (VU' - UV ') / V ²

If X-1 under the root minus 1-x under the root plus four equals the square of X + y, find the square of X + the square of Y - 2XY

In order to make "X-1 under the root sign" and "1-x under the root sign" meaningful, we know that x = 1
Further, it is obtained that 4 = (1 + y) 2, that is, y = 1 or - 3
"Square of X + square of Y - 2XY" = (X-Y) 2 = 0 or 16

If y is equal to X-2 / 2, the square of the root sign x minus 4 plus the root sign 4 minus the square of X, and then add 3 to find the value of the Y square of X

X in root sign ²- 4>=0,x ²>= four
4-x ²>= 0,x ²

If f '(x) = x and f (0) = 0, then ∫ f (x) DX

By F '(x) = x, f (0) = 0,
Get f (x) = 1 / 2x ^ 2
∫f(x)dx=1/6x^3+c
Hope to adopt