Step by step solution to "Find the four second partial derivatives of f(x,y)=x2y3."

Reggie

Answered question

2021-03-07

Find the four second partial derivatives of

f (x, y) = x^{2} y^{3}

Answer & Explanation

Clelioo

Skilled2021-03-08Added 88 answers

Step 1
Given function:
$f (x, y) = x^{2} y^{3}$
The partial derivatives over four seconds of $f (x, y) : f_{x x}, f_{y y}, f_{x y}, f_{y x}$
Step 2
Now,
Apply the Power Rule: $\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}$ Identifying the partial derivatives
$f_{x x} = \frac{\partial}{\partial x} (\frac{\partial f}{\partial x})$
$= \frac{\partial}{\partial x} (\frac{\partial x^{2} y^{3}}{\partial x})$
$= (\partial \frac{2 x y^{3}}{\partial x} = 2 y^{3}$
$f_{y y} = \frac{\partial}{\partial y} (\frac{\partial f}{\partial y})$
$= \frac{\partial}{\partial y} (\frac{\partial x^{2} y^{3}}{\partial} y))$
$= (\partial \frac{3 x^{2} y^{2}}{\partial} y)$
$= 3 x^{2} \times 2 y$
$= 6 x^{2} y$
$f_{x y} = \frac{\partial}{\partial y} (\frac{\partial f}{\partial x})$
$= \frac{\partial}{\partial y} \frac{\partial x^{2} y^{3}}{\partial x})$
$= (\partial \frac{2 x y^{3}}{\partial y})$
$= 6 x^{2}$
$f_{y x} = \frac{\partial}{\partial x} (\frac{\partial f}{\partial y})$
$= \frac{\partial}{\partial x} (\frac{\partial x^{2} y^{3}}{\partial y})$
$= (\partial \frac{3 x^{2} y^{2}}{\partial x})$
$= 6 x y^{2}$

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