Question

Use the Chain Rule to find ∂z/∂s and ∂z/∂t.

z = eˣ⁺⁷ʸ, x = s2t, y = st2
∂z/∂s = _____
∂z/∂t = ______

Answer 1

To find \(\partial z/\partial s\) and \(\partial z/\partial t\), apply the Chain Rule by taking partial derivatives of z with respect to x and y, and then use the derivatives of x and y with respect to s and t.

Step-by-step explanation:

To solve for \(\partial z/\partial s\) and \(\partial z/\partial t\) using the Chain Rule, we will take the partial derivatives of z with respect to x and y, and then use the derivatives of x and y with respect to s and t.

Given z = e^{x+7y}, x = s^2t, and y = st^2, we first compute the partial derivatives of z with respect to x and y:

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• \(\frac{\partial z}{\partial x} = e^{x+7y}\)
• 
  
• \(\frac{\partial z}{\partial y} = 7e^{x+7y}\)

Now we find the partial derivatives of x and y with respect to s and t:

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• \(\frac{\partial x}{\partial s} = 2st\)
• 
  
• \(\frac{\partial x}{\partial t} = s^2\)
• 
  
• \(\frac{\partial y}{\partial s} = t^2\)
• 
  
• \(\frac{\partial y}{\partial t} = 2st\)

Then, using the Chain Rule, we can find:

1. 
   
3. \(\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s} = e^{x+7y}\cdot2st + 7e^{x+7y}\cdot t^2\)
4. 
   
6. \(\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t} = e^{x+7y}\cdot s^2 + 7e^{x+7y}\cdot 2st\)

Therefore, the partial derivatives of z with respect to s and t are:

• 
  
• \(\partial z/\partial s = e^{x+7y}\cdot2st + 7e^{x+7y}\cdot t^2\)
• 
  
• \(\partial z/\partial t = e^{x+7y}\cdot s^2 + 7e^{x+7y}\cdot 2st\)