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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).

User Divanov
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Answer:

Explanation:

Given that


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

Let us find partial derivatives one by one


(∂f )/(∂x) =1\\(∂^2f )/(∂y∂x) =(∂ )/(∂y)(1) =0\\(∂^3f )/(∂z∂y∂x) =(∂ )/(∂z)(0) =0

At the point (2,1,-2)


(∂^2f )/(∂y∂x)=0

(since at all points the value is constant 0)

User Loaf
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