Math Problem Statement

Given the function f (y, z) = (sin y)ez3−z−1

  • eyz (a) Determine fz (b) Determine fy (c) Determine fyz

Solution

Given the function f(y,z)=(siny)ez3z1+eyzf(y, z) = (\sin y) e^{z^3 - z - 1} + e^{yz}, we will compute each of the partial derivatives step by step.

(a) Determine fzf_z (the partial derivative with respect to zz):

We apply the rules of differentiation, treating yy as a constant.

First term:

(siny)ez3z1(\sin y) e^{z^3 - z - 1}

The derivative of ez3z1e^{z^3 - z - 1} with respect to zz is: ddz(ez3z1)=ez3z1ddz(z3z1)=ez3z1(3z21)\frac{d}{dz}\left(e^{z^3 - z - 1}\right) = e^{z^3 - z - 1} \cdot \frac{d}{dz}(z^3 - z - 1) = e^{z^3 - z - 1} \cdot (3z^2 - 1)

Thus, the derivative of the first term is: (siny)ez3z1(3z21)(\sin y) \cdot e^{z^3 - z - 1} \cdot (3z^2 - 1)

Second term:

eyze^{yz}

The derivative of eyze^{yz} with respect to zz is: ddz(eyz)=yeyz\frac{d}{dz}(e^{yz}) = y e^{yz}

So the partial derivative fzf_z is: fz=(siny)ez3z1(3z21)+yeyzf_z = (\sin y) e^{z^3 - z - 1} (3z^2 - 1) + y e^{yz}

(b) Determine fyf_y (the partial derivative with respect to yy):

We now differentiate with respect to yy, treating zz as constant.

First term:

(siny)ez3z1(\sin y) e^{z^3 - z - 1}

The derivative of siny\sin y is cosy\cos y. So, the derivative of the first term is: (cosy)ez3z1(\cos y) e^{z^3 - z - 1}

Second term:

eyze^{yz}

The derivative of eyze^{yz} with respect to yy is: ddy(eyz)=zeyz\frac{d}{dy}(e^{yz}) = z e^{yz}

So the partial derivative fyf_y is: fy=(cosy)ez3z1+zeyzf_y = (\cos y) e^{z^3 - z - 1} + z e^{yz}

(c) Determine fyzf_{yz} (the mixed partial derivative with respect to yy and zz):

Now, we differentiate fyf_y with respect to zz.

From part (b), we know that: fy=(cosy)ez3z1+zeyzf_y = (\cos y) e^{z^3 - z - 1} + z e^{yz}

First term:

(cosy)ez3z1(\cos y) e^{z^3 - z - 1}

The derivative with respect to zz is: (cosy)ez3z1(3z21)(\cos y) \cdot e^{z^3 - z - 1} \cdot (3z^2 - 1)

Second term:

zeyzz e^{yz}

The derivative with respect to zz is: ddz(zeyz)=eyz+zyeyz\frac{d}{dz}(z e^{yz}) = e^{yz} + z y e^{yz}

So the mixed partial derivative fyzf_{yz} is: fyz=(cosy)ez3z1(3z21)+eyz+zyeyzf_{yz} = (\cos y) e^{z^3 - z - 1} (3z^2 - 1) + e^{yz} + z y e^{yz}

Summary of Results:

(a) fz=(siny)ez3z1(3z21)+yeyzf_z = (\sin y) e^{z^3 - z - 1} (3z^2 - 1) + y e^{yz}

(b) fy=(cosy)ez3z1+zeyzf_y = (\cos y) e^{z^3 - z - 1} + z e^{yz}

(c) fyz=(cosy)ez3z1(3z21)+eyz+zyeyzf_{yz} = (\cos y) e^{z^3 - z - 1} (3z^2 - 1) + e^{yz} + z y e^{yz}

Would you like more details or have any questions?


Here are 5 related questions to expand on this topic:

  1. What is the second partial derivative fzzf_{zz} of the function?
  2. How does the chain rule apply when computing partial derivatives?
  3. Can you find the critical points of the function f(y,z)f(y, z)?
  4. How would the function change if the exponent in eyze^{yz} were y2z2y^2 z^2?
  5. What happens if y=0y = 0 or z=0z = 0? How does this simplify f(y,z)f(y, z)?

Tip: When differentiating exponentials with a variable in the exponent, remember to apply the chain rule carefully!

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Math Problem Analysis

Mathematical Concepts

Partial Derivatives
Multivariable Calculus
Exponentiation
Trigonometric Functions

Formulas

f_z = (\sin y) e^{z^3 - z - 1} (3z^2 - 1) + y e^{yz}
f_y = (\cos y) e^{z^3 - z - 1} + z e^{yz}
f_{yz} = (\cos y) e^{z^3 - z - 1} (3z^2 - 1) + e^{yz} + z y e^{yz}

Theorems

Chain Rule
Product Rule

Suitable Grade Level

College Level (Calculus II or Multivariable Calculus)