Answer:
[tex]\dfrac{\partial z}{\partial x}=6e^{x^3-y^3}x^5+6e^{x^3-y^3}x^2[/tex]
[tex]\dfrac{\partial z}{\partial y}=-6e^{x^3-y^3}x^3y^2[/tex]
Step-by-step explanation:
Partial derivatives are the rates at which a multivariable function changes with respect to specific individual variables, while keeping all other variables constant.
Given:
[tex]z=(u+v)e^v\\\\u=x^3+y^3\\\\v=x^3-y^3[/tex]
To find δz/δx and δz/δy, begin by writing z in terms of x and y by substituting the given expressions for u and v into the equation for z:
[tex]z=(x^3+y^3+x^3-y^3)e^{x^3-y^3}\\\\z=2x^3e^{x^3-y^3}[/tex]
[tex]\dotfill[/tex]
To find the partial derivative δz/δx, we differentiate the terms in x with respect to x, while treating the term in y as a constant.
For this, we can use the product rule.
[tex]\boxed{\begin{array}{c}\underline{\textsf{Product Rule}}\\\\\dfrac{\text{d}}{\text{d}x}\left[f(x) \cdot g(x)\right]=f'(x) \cdot g(x) + f(x) \cdot g'(x)\end{array}}[/tex]
In this case:
[tex]f(x) = 2x^3 \implies f'(x)=6x^2[/tex]
[tex]g(x) = e^{x^3-y^3}\implies g'(x)=3x^2e^{x^3-y^3}[/tex]
Plug everything into the product rule:
[tex]\dfrac{\partial z}{\partial x}=6x^2 \cdot e^{x^3-y^3}+2x^3\cdot 3x^2e^{x^3-y^3}[/tex]
Simplify:
[tex]\dfrac{\partial z}{\partial x}=6x^2 e^{x^3-y^3}+6x^{3+2}e^{x^3-y^3}[/tex]
[tex]\dfrac{\partial z}{\partial x}=6x^2e^{x^3-y^3}+6x^5e^{x^3-y^3}[/tex]
[tex]\dfrac{\partial z}{\partial x}=6e^{x^3-y^3}x^5+6e^{x^3-y^3}x^2[/tex]
Therefore, the partial derivative δz/δx is:
[tex]\Large\boxed{\boxed{\dfrac{\partial z}{\partial x}=6e^{x^3-y^3}x^5+6e^{x^3-y^3}x^2}}[/tex]
[tex]\dotfill[/tex]
To find the partial derivative δz/δy, we differentiate the term in y with respect to y, while treating the terms in x as constants.
[tex]\dfrac{\partial z}{\partial y}=\dfrac{\partial}{\partial y}\left(2x^3e^{x^3-y^3}\right)[/tex]
Take the constant 2x³ out:
[tex]\dfrac{\partial z}{\partial y}=2x^3\cdot \dfrac{\partial}{\partial y}\left(e^{x^3-y^3}\right)[/tex]
Use the chain rule to differentiate [tex]e^{x^3-y^3}[/tex], which means we multiply [tex]e^{x^3-y^3}[/tex] by the derivative of its exponent:
[tex]\dfrac{\partial z}{\partial y}=2x^3 \cdot e^{x^3-y^3} \cdot \dfrac{\partial}{\partial y}\left(x^3-y^3\right)[/tex]
Since x³ is a constant, it becomes zero:
[tex]\dfrac{\partial z}{\partial y}=2x^3 \cdot e^{x^3-y^3} \cdot -3y^2[/tex]
Simplify:
[tex]\dfrac{\partial z}{\partial y}=-6e^{x^3-y^3}x^3y^2[/tex]
Therefore, the partial derivative δz/δy is:
[tex]\Large\boxed{\boxed{\dfrac{\partial z}{\partial y}=-6e^{x^3-y^3}x^3y^2}}[/tex]
