Question

D^2z/dxdy, где z=x^3+4xy+7y-1

Answer 1

$\frac{d^2 z}{dxdy}= \frac{d}{dx} (\frac{dz}{d y})\\ \frac{dz}{d y} = 4x+7 \\ \frac{d^2 z}{dxdy}= \frac{d}{dx} (4x+7)=4$
Ответ: 4.

Answer 2

To find the second mixed partial derivative of the function z with respect to x and y, denoted as $\frac{{d^2z}}{{dxdy}}$, we need to find the first mixed partial derivative with respect to x and then differentiate the result with respect to y. Let's go through the steps:

Given: $z = x^3 + 4xy + 7y - 1$

Step 1: Find the first partial derivative of z with respect to x, denoted as $\frac{{dz}}{{dx}}$: $\frac{{dz}}{{dx}} = \frac{{d}}{{dx}}(x^3 + 4xy + 7y - 1)$

Now, we'll differentiate each term with respect to x: $\frac{{d}}{{dx}}(x^3) = 3x^2$ $\frac{{d}}{{dx}}(4xy) = 4y$ (treat y as a constant) $\frac{{d}}{{dx}}(7y) = 0$ (constant with respect to x) $\frac{{d}}{{dx}}(-1) = 0$ (constant with respect to x)

So, $\frac{{dz}}{{dx}} = 3x^2 + 4y$

Step 2: Now, find the second partial derivative of $\frac{{dz}}{{dx}}$ with respect to y, denoted as $\frac{{d^2z}}{{dxdy}}$: $\frac{{d^2z}}{{dxdy}} = \frac{{d}}{{dy}}\left(\frac{{dz}}{{dx}}\right)$

Now, differentiate $\frac{{dz}}{{dx}}$ with respect to y: $\frac{{d}}{{dy}}(3x^2 + 4y) = 4$

So, the second mixed partial derivative $\frac{{d^2z}}{{dxdy}} = 4$.