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$4-2025$ VE

Math

Question $25,$

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Consider the functions $z=-9e^{x}lny,x=ln(ucosv),$ and $y=$usinv.

$($a$)$ Express $\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{u}}$ and $\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{v}}$ as functions of $u$ and $v$ both by using the Chain Rule and by expressing $z$ directly in terms of $u$ and $v$ before differentiating.

$($b$)$ Evaluate $\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{u}}$ and $\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{v}}$ at $(u,v)=(2,\frac{}{4}).$

$($a$)$ Find each partial derivative needed to use the Chain Rule to find $\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{u}}.$

$\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{x}}=$

$\frac{\text{d}\text{e}\text{l}\text{x}}{\text{d}\text{e}\text{l}\text{u}}=$

$\frac{\text{d}\text{e}\text{l}\text{y}}{\text{d}\text{e}\text{l}\text{u}}=$

$\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{y}}=$