Find the following derivatives.

z Subscript szs 

and 

z Subscript tzt​, 

where 

zequals=2 xy minus 2 x squared y2xy−2x2y​, 

xequals=2 s plus 3 t2s+3t​, 

and 

yequals=2 s minus 3 t

To find the derivatives \( z_{s} \) and \( z_{t} \), we can apply the **multivariable chain rule** since \( z \) is expressed as a function of both \( x \) and \( y \), which in turn are functions of \( s \) and \( t \).

### Given:
\[
z = 2xy - 2x^2y
\]
\[
x = 2s + 3t
\]
\[
y = 2s - 3t
\]

### 1. Finding \( z_s \) (the partial derivative of \( z \) with respect to \( s \)):

By the chain rule:
\[
z_s = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s}
\]

First, we compute the partial derivatives:
- \( \frac{\partial z}{\partial x} = 2y - 4xy \)
- \( \frac{\partial z}{\partial y} = 2x - 2x^2 \)
- \( \frac{\partial x}{\partial s} = 2 \)
- \( \frac{\partial y}{\partial s} = 2 \)

Now, substitute these into the equation for \( z_s \):
\[
z_s = (2y - 4xy)(2) + (2x - 2x^2)(2)
\]
\[
z_s = 2(2y - 4xy) + 2(2x - 2x^2)
\]
\[
z_s = 4y - 8xy + 4x - 4x^2
\]

### 2. Finding \( z_t \) (the partial derivative of \( z \) with respect to \( t \)):

Similarly, by the chain rule:
\[
z_t = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t}
\]

We already know:
- \( \frac{\partial z}{\partial x} = 2y - 4xy \)
- \( \frac{\partial z}{\partial y} = 2x - 2x^2 \)
- \( \frac{\partial x}{\partial t} = 3 \)
- \( \frac{\partial y}{\partial t} = -3 \)

Substitute these into the equation for \( z_t \):
\[
z_t = (2y - 4xy)(3) + (2x - 2x^2)(-3)
\]
\[
z_t = 3(2y - 4xy) - 3(2x - 2x^2)
\]
\[
z_t = 6y - 12xy - 6x + 6x^2
\]

### Final answers:

- \( z_s = 4y - 8xy + 4x - 4x^2 \)
- \( z_t = 6y - 12xy - 6x + 6x^2 \)

Would you like more details or have any questions?

Here are 5 follow-up questions:
1. Can you explain the chain rule in more depth?
2. How would these partial derivatives change if \( z \) were expressed differently?
3. Can we generalize this process to three variables?
4. What is the significance of mixed partial derivatives in multivariable calculus?
5. How would you solve for the critical points of this function?

**Tip**: Remember that in chain rule applications, it's crucial to carefully track each partial derivative of intermediate variables.

Find the following derivatives: z_s and z_t, where z = 2xy - 2x^2y, x = 2s + 3t, and y = 2s - 3t.

Learn how to compute the partial derivatives of z with respect to s and t, where z = 2xy - 2x^2y, x = 2s + 3t, and y = 2s - 3t. This problem involves the multivariable chain rule and partial derivatives, suitable for undergraduate mathematics students.

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