Question

Let f(x, y, z) = x + ln(2y + 3z 2 ). Compute ∂f ∂x, ∂ 2f ∂y∂x, and ∂ 3f ∂z∂y∂x. Evaluate ∂ 2f ∂y∂x(2, 1, −2).

Answer 1

Answer:

Step-by-step explanation:

Given that

$f(x, y, z) = x + ln(2y + 3z^2 )$

Let us find partial derivatives one by one

$\frac{∂f }{∂x} =1\\\frac{∂^2f }{∂y∂x} =\frac{∂ }{∂y}(1) =0\\\frac{∂^3f }{∂z∂y∂x} =\frac{∂ }{∂z}(0) =0$

At the point (2,1,-2)

$\frac{∂^2f }{∂y∂x}=0$

(since at all points the value is constant 0)