Ah, no! Let z = f (Y / x), where f (U) is a differentiable function, and prove that: X（ α z/ α x)+y( α z/ α y)=0 Ask for expert advice!

Ah, no! Let z = f (Y / x), where f (U) is a differentiable function, and prove that: X（ α z/ α x)+y( α z/ α y)=0 Ask for expert advice!

If u = Y / x, then（ α u/ α x)=(-y/x^2),( α u/ α y)=(1/x)
For Z = f (Y / x) = f (U), there are
( α z/ α x)=( α z/ α u)*( α u/ α x)=( α z/ α u)(-y/x^2),x( α z/ α x)=x*( α z/ α u)(-y/x^2)=-( α z/ α u)(y/x)
( α z/ α y)=( α z/ α u)*( α u/ α y)=( α z/ α u)(1/x),y( α z/ α y)=y*( α z/ α u)(1/x)=( α z/ α u)(y/x)
∴x( α z/ α x)+y( α z/ α y)=-( α z/ α u)(y/x)+( α z/ α u)(y/x)=0

Let f (U, V) be a differentiable function, and the equation f (x + Z / y, y + Z / x) = 0. The determined function z = (x, y) proves X*（ α z/ α x)+y*( α Z/ α y)=z-xy α Partial derivative

Let the partial derivatives of F with respect to u and V be written as f'1 and f'2 respectively, and write that f'1 (x + Z / y) = a, f'2 (y + Z / x) = B (both a and B are expressions about X, y, z). Then, f (x + Z / y, y + Z / x) = 0 is derived by the partial derivation rule of composite function α F/ α x=a -bz/(x^2) α F/ α y= - az/(y^2)+b α F/ α Z = A / y + B / x, so x*（ α z/...

Z = Y / F (x ^ 2-y ^ 2), where f (U) is differentiable, verification

This is the use of the derivative of a composite function
δ z/ δ x =2xyf'/f ²
δ z/ δ y =[f+yf'(-2y)]/f ²= (f-2y ² f')/f ²
1/x ×δ z/ δ x+1/y ×δ z/ δ y
=2yf'/f ²+ 1/yf-2yf'/f ²
=1/yf
=z/y ²

Function y = e to the power of X, the derivative of this function on x0 is 2, find x0 =?, Although the answer is in2, I want to know the process!

y=e^x
y'=e^x
When y '= e ^ x0 = 2
x0=ln2

Is the function f (x) = e ^| x-x0| differentiable at x = x0

When x = x0 +, f (x) = e ^ (x-x0), f '(x) = e ^ (x-x0), f' (x0 +) = 1
When x = x0 -, f (x) = e ^ (x0-x), f '(x) = - e ^ (x0-x), f' (x0 -) = - 1
f'(x0+)f'(x0-)
Therefore, it is not differentiable at x = x0

If the function y = f (x) gets the maximum value at point x = x0, there must be () answer f '(x0) = 0 or does not exist X0 is a representation, 0 is not a power

If the function is differentiable at x0, the derivative is 0 and the maximum value is taken
If it is not differentiable, that is, the derivative does not exist, it is also possible to take the maximum value and consider the absolute value of the function y = X
There is no function y = 1x1 without process proof
The maximum value is removed at point 0, but the left derivative and right derivative are not equal, and the limit does not exist