Find the following derivatives. \( z_{s} \) and \( z_{\text {, }} \), where \( z=7 x y-2 x^{2} y, x=2 s+5 t \), and \( y=2 s-5 t \) \( \frac{\sigma z}{\partial y}=7 x-2 x^{2} \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{\partial y}{\partial s}=2 \) (Type an expression using s and \( t \) as the variables.) \( \frac{\partial x}{\partial t}=5 \) (Type an expression using \( s \) and \( t \) as the variables.) \( \frac{\partial y}{\partial t}=-5 \) (Type an expression using \( s \) and \( t \) as the variables.) \( z_{s}=-48 s^{2}-80 s t+56 s+100 t^{2} \) (Type an expression using \( s \) and \( t \) as the variables.) \( z_{4}=\square \) (Type an expression using \( s \) and \( t \) as the variables.)

To find the derivatives \( z_s \) and \( z_{,} \) where \( z = 7xy - 2x^2y \), we will use the chain rule and the given relationships for \( x \) and \( y \). ### Step 1: Find \( z_s \) Using the chain rule, we have: \[ z_s = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s} \] #### Step 1.1: Calculate \( \frac{\partial z}{\partial x} \) Given \( z = 7xy - 2x^2y \): \[ \frac{\partial z}{\partial x} = 7y - 4xy \] #### Step 1.2: Calculate \( \frac{\partial z}{\partial y} \) From the problem statement: \[ \frac{\partial z}{\partial y} = 7x - 2x^2 \] #### Step 1.3: Calculate \( \frac{\partial x}{\partial s} \) and \( \frac{\partial y}{\partial s} \) From the relationships: \[ x = 2s + 5t \implies \frac{\partial x}{\partial s} = 2 \] \[ y = 2s - 5t \implies \frac{\partial y}{\partial s} = 2 \] ### Step 2: Substitute into the equation for \( z_s \) Now substituting everything into the equation for \( z_s \): \[ z_s = (7y - 4xy) \cdot 2 + (7x - 2x^2) \cdot 2 \] ### Step 3: Simplify \( z_s \) \[ z_s = 2(7y - 4xy) + 2(7x - 2x^2) \] \[ = 14y - 8xy + 14x - 4x^2 \] ### Step 4: Find \( z_{,} \) To find \( z_{,} \), we will use the chain rule again: \[ z_{,} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t} \] #### Step 4.1: Calculate \( \frac{\partial x}{\partial t} \) and \( \frac{\partial y}{\partial t} \) From the relationships: \[ \frac{\partial x}{\partial t} = 5 \] \[ \frac{\partial y}{\partial t} = -5 \] ### Step 5: Substitute into the equation for \( z_{,} \) Now substituting into the equation for \( z_{,} \): \[ z_{,} = (7y - 4xy) \cdot 5 + (7x - 2x^2) \cdot (-5) \] ### Step 6: Simplify \( z_{,} \) \[ z_{,} = 5(7y - 4xy) - 5(7x - 2x^2) \] \[ = 35y - 20xy - 35x + 10x^2 \] ### Final Expressions 1. \( z_s = 14y - 8xy + 14x - 4x^2 \) 2. \( z_{,} = 35y - 20xy - 35x + 10x^2 \) Now, let's summarize the expressions for \( z_s \) and \( z_{,} \): - \( z_s = 14y - 8xy + 14x - 4x^2 \) - \( z_{,} = 35y - 20xy - 35x + 10x^2 \) ### Additional Information For the expressions involving \( z_4 \), we need to evaluate \( z \) at \( s = 4 \) and substitute the values of \( x \) and \( y \) accordingly. 1. Calculate \( x \) and \( y \) at \( s = 4 \): - \( x = 2(4) + 5t = 8 + 5t \) - \( y = 2(4) - 5t = 8 - 5t \) 2. Substitute \( x \) and \( y \) into \( z \): \[ z = 7(8 + 5t)(8 - 5t) - 2(8 + 5t)^2(8 - 5t) \] This expression can be simplified further to find \( z_4 \). Would you like me to perform that calculation?