Question

Obtain all the first and second order partial derivatives of the function: f (x, y)=x^2 sin y+y^2 cos x

Answer 1

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Answer on Question # 72163 – Math – Differential Equations

Question

Obtain all the first and second order partial derivatives of the function: $f(x, y) = x^2 \sin y + y^2 \cos x$

Solution

We have the function of two variables

$$f(x, y) = x^2 \sin y + y^2 \cos x$$

1) Obtain the first order partial derivatives.

Holding $y$ constant and differentiating $f$ with respect to $x$, we get $f_x$:

$$f_x = (x^2 \sin y + y^2 \cos x)_x = 2x \sin y - y^2 \sin x$$

Holding $x$ constant and differentiating $f$ with respect to $y$, we get $f_y$:

$$f_y = (x^2 \sin y + y^2 \cos x)_y = x^2 \cos y + 2y \cos x$$

2) Obtain the second order partial derivatives.

Holding $y$ constant and differentiating $f_x$ with respect to $x$, we get $f_{xx}$:

$$f_{xx} = (f_x)_x = (2x \sin y - y^2 \sin x)_x = 2 \sin y - y^2 \cos x$$

Holding $x$ constant and differentiating $f_x$ with respect to $y$, we get $f_{xy}$:

$$f_{xy} = (f_x)_y = (2x \sin y - y^2 \sin x)_y = 2x \cos y - 2y \sin x$$

Holding $y$ constant and differentiating $f_y$ with respect to $x$, we get $f_{yx}$:

$$f_{yx} = (f_y)_x = (x^2 \cos y + 2y \cos x)_x = 2x \cos y - 2y \sin x$$

Thus $f_{xy} = f_{yx}$

Holding $x$ constant and differentiating $f_y$ with respect to $y$, we get $f_{yy}$:

$$f_{yy} = (f_y)_y = (x^2 \cos y + 2y \cos x)_y = -x^2 \sin y + 2 \cos x$$

Answer:

The first order partial derivatives are

$$f_x = 2x \sin y - y^2 \sin x$$ $$f_y = x^2 \cos y + 2y \cos x$$

The second order partial derivatives are

$$f_{xx} = 2 \sin y - y^2 \cos x$$ $$f_{xy} = f_{yx} = 2x \cos y - 2y \sin x$$ $$f_{yy} = -x^2 \sin y + 2 \cos x$$