Find the second partial derivative of Z = ln (x * x + y * y)

Find the second partial derivative of Z = ln (x * x + y * y)

∂z/∂x=2x/(x ²+ y ²) ∂z/∂y=2y/(x ²+ y ²) ∂ ² z/∂x∂y=[2(x ²+ y ²)- 2x*2y]/(x ²+ y ²)²= 2(x-y) ²/ (x ²+ y ²)² W...

Given z = ln (x, y), find the first and second partial derivatives of Z,

Given z = ln (x, y), find the first and second partial derivatives of Z
∂z/∂x=y/xy=1/x； ∂z/∂y=x/xy=1/y；
∂ ² z/∂x ²=- 1/x ²； ∂ ² z/∂y ²=- 1/y ²；
∂ ² z/∂x∂y=0； ∂ ² z/∂y∂x=0；

Find the second derivative arctan X / y = ln radical x ^ 2 + y ^ 2

Important steps for writing directly:
Take the derivative of X at both ends, simplify it, and get
y-y'x=2x+2y-y'
y'=(y-2x)/(x+2y)
Then take the derivative of X at both ends, simplify it, and substitute the result of the previous step into it to obtain
y''=-10(x^2+y^2)/ (x+2y)^3

1. Find the derivative of y = ln (x + 1). 2. Find the derivative of e ^ 2x Let t = x + 1, then the derivative is 1 / (x + 1) Let t = 2x, then the derivative is e ^ 2x Why is 1 right and 2 wrong

The second question is wrong because you don't understand the derivation of composite functions. I suggest you take a look at the derivation of composite functions
The positive solution is as follows. Let t = 2x f (x) = e ^ 2x = e ^ t
f'(x) =f'(t)*t‘（x)
=(e^t)*2
Then replace T with X and push out f '(x) = 2E ^ (2x)

Find the derivative of y = x-ln (2e ^ x + 1 + √ (e ^ 2x + 4E ^ x + 1)) How

y=1-1/(2e^x+1+√(e^2x+4e^x+1))*(2e^x+1/2*((e^2x+4e^x+1))^(-1/2)*(2e^(2x)+4e^x)))

The function f (x) = sin2x, G (x) = cos (2x + π) is known 6) , the straight line x = t (t ∈ R). It intersects with the images of functions f (x) and G (x) at two points m and N, respectively (1) When t = π 4, find the value of | Mn |; (2) Find Mn in t ∈ [0, π 2] Maximum value at

(1) Substituting t = π 4 into functions f (x) and G (x), we get ∵ Mn | = f (π 4) − g (π 4) | = | sin (2) × π4)−cos(2 × π4+π6)|=|1−cos2π3|＝32.（2）∵|MN|＝|f(t)−g(t)|＝|sin2t−cos(2t+π6)|=|32sin2t−32cos2t|=3|sin(2t−π6)|∵t...