To find the gradient of the function \( v = e^{-z} \sin(2x) \cosh(y) \), we need to find the partial derivatives of \( v \) with respect to \( x \), \( y \), and \( z \).

### Partial Derivative with respect to \( x \)

\[ \frac{\partial v}{\partial x} = \frac{\partial}{\partial x} \left( e^{-z} \sin(2x) \cosh(y) \right) \]

Since \( e^{-z} \) and \( \cosh(y) \) are constants with respect to \( x \), we get:

\[ \frac{\partial v}{\partial x} = e^{-z} \cosh(y) \frac{\partial}{\partial x} (\sin(2x)) \]

The derivative of \( \sin(2x) \) is \( 2 \cos(2x) \), so:

\[ \frac{\partial v}{\partial x} = e^{-z} \cosh(y) \cdot 2 \cos(2x) \]
\[ \frac{\partial v}{\partial x} = 2 e^{-z} \cosh(y) \cos(2x) \]

### Partial Derivative with respect to \( y \)

\[ \frac{\partial v}{\partial y} = \frac{\partial}{\partial y} \left( e^{-z} \sin(2x) \cosh(y) \right) \]

Since \( e^{-z} \) and \( \sin(2x) \) are constants with respect to \( y \), we get:

\[ \frac{\partial v}{\partial y} = e^{-z} \sin(2x) \frac{\partial}{\partial y} (\cosh(y)) \]

The derivative of \( \cosh(y) \) is \( \sinh(y) \), so:

\[ \frac{\partial v}{\partial y} = e^{-z} \sin(2x) \cdot \sinh(y) \]
\[ \frac{\partial v}{\partial y} = e^{-z} \sin(2x) \sinh(y) \]

### Partial Derivative with respect to \( z \)

\[ \frac{\partial v}{\partial z} = \frac{\partial}{\partial z} \left( e^{-z} \sin(2x) \cosh(y) \right) \]

Since \( \sin(2x) \) and \( \cosh(y) \) are constants with respect to \( z \), we get:

\[ \frac{\partial v}{\partial z} = \sin(2x) \cosh(y) \frac{\partial}{\partial z} (e^{-z}) \]

The derivative of \( e^{-z} \) is \( -e^{-z} \), so:

\[ \frac{\partial v}{\partial z} = \sin(2x) \cosh(y) \cdot (-e^{-z}) \]
\[ \frac{\partial v}{\partial z} = -\sin(2x) \cosh(y) e^{-z} \]

### Summary

The gradient of \( v = e^{-z} \sin(2x) \cosh(y) \) is:

\[ \nabla v = \left( \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}, \frac{\partial v}{\partial z} \right) \]

\[ \nabla v = \left( 2 e^{-z} \cosh(y) \cos(2x), e^{-z} \sin(2x) \sinh(y), -\sin(2x) \cosh(y) e^{-z} \right) \]

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### Related Questions

1. How do you find the second partial derivatives of \( v = e^{-z} \sin(2x) \cosh(y) \)?
2. What is the gradient of \( v = x^2y \ln(z) \)?
3. How do you find the gradient of \( v = x^3 + y^3 + z^3 \)?
4. What is the partial derivative of \( v = e^{x+y+z} \) with respect to \( y \)?
5. How do you compute the gradient of \( v = \cos(xy) \sin(z) \)?
6. What is the gradient of \( v = x e^{yz} \)?
7. How do you find the gradient of \( v = \ln(x^2 + y^2 + z^2) \)?
8. What is the gradient of \( v = \frac{xy}{z} \)?

### Tip

When differentiating exponential functions with negative exponents, remember that the derivative of \( e^{-x} \) is \( -e^{-x} \).

Learn how to find the gradient of the function v = e^-z sin(2x) cosh(y) using partial derivatives. This problem involves exponential, trigonometric, and hyperbolic functions. Suitable for college-level mathematics.

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