Find [tex]$\frac{\partial z}{\partial x}$[/tex] and [tex]$\frac{\partial z}{\partial y}$[/tex] for the function [tex]$x^2 \sin \left(y^3\right) + x e^{3z} - \cos \left(z^2\right) = 3y - 6z + 8$[/tex].



Answer :

To find the partial derivatives [tex]\(\frac{\partial z}{\partial x}\)[/tex] and [tex]\(\frac{\partial z}{\partial y}\)[/tex] for the function given by:

[tex]\[ f(x, y, z) = x^2 \sin(y^3) + x e^{3z} - \cos(z^2) - 3y + 6z - 8 = 0 \][/tex]

we need to compute the partial derivatives of this implicit function with respect to [tex]\(x\)[/tex] and [tex]\(y\)[/tex].

### Step-by-Step Solution

1. Given Function:
[tex]\[ f(x, y, z) = x^2 \sin(y^3) + x e^{3z} - \cos(z^2) - 3y + 6z - 8 \][/tex]

2. Find [tex]\(\frac{\partial f}{\partial x}\)[/tex]:
[tex]\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( x^2 \sin(y^3) + x e^{3z} - \cos(z^2) - 3y + 6z - 8 \right) \][/tex]

Breaking it down:
- [tex]\(\frac{\partial}{\partial x} \left(x^2 \sin(y^3)\right) = 2x \sin(y^3) \)[/tex]
- [tex]\(\frac{\partial}{\partial x} \left(x e^{3z}\right) = e^{3z} \)[/tex]
- [tex]\(\frac{\partial}{\partial x} \left(- \cos(z^2)\right) = 0 \)[/tex]
- [tex]\(\frac{\partial}{\partial x} \left(-3y\right) = 0 \)[/tex]
- [tex]\(\frac{\partial}{\partial x} \left(6z\right) = 0 \)[/tex]
- [tex]\(\frac{\partial}{\partial x} \left(- 8\right) = 0 \)[/tex]

Combining these results:

[tex]\[ \frac{\partial f}{\partial x} = 2x \sin(y^3) + e^{3z} \][/tex]

3. Find [tex]\(\frac{\partial f}{\partial y}\)[/tex]:
[tex]\[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( x^2 \sin(y^3) + x e^{3z} - \cos(z^2) - 3y + 6z - 8 \right) \][/tex]

Breaking it down:
- [tex]\(\frac{\partial}{\partial y} \left(x^2 \sin(y^3)\right) = x^2 \frac{\partial}{\partial y} \left( \sin(y^3) \right) = x^2 \left(3y^2 \cos(y^3)\right) = 3x^2 y^2 \cos(y^3)\)[/tex]
- [tex]\(\frac{\partial}{\partial y} \left(x e^{3z}\right) = 0 \)[/tex]
- [tex]\(\frac{\partial}{\partial y} \left(- \cos(z^2)\right) = 0 \)[/tex]
- [tex]\(\frac{\partial}{\partial y} \left(-3y\right) = -3 \)[/tex]
- [tex]\(\frac{\partial}{\partial y} \left(6z\right) = 0 \)[/tex]
- [tex]\(\frac{\partial}{\partial y} \left(- 8\right) = 0 \)[/tex]

Combining these results:

[tex]\[ \frac{\partial f}{\partial y} = 3x^2 y^2 \cos(y^3) - 3 \][/tex]

Therefore, the computed partial derivatives are:
[tex]\[ \frac{\partial z}{\partial x} = 2x \sin(y^3) + e^{3z} \][/tex]
[tex]\[ \frac{\partial z}{\partial y} = 3x^2 y^2 \cos(y^3) - 3 \][/tex]