Question

Ache as derivadas parciais primeiras da função $f(p,V) = ln \sqrt{\frac{p + V}{p - V}$ .

Answer 1

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$\sf f(p,v)=\ell n\sqrt{\dfrac{p+v}{p-v}}\\\sf f(p,v)=\ell n\left(\dfrac{p+v}{p-v}\right)^{\frac{1}{2}}\\\sf f(p,v)=\dfrac{1}{2}\ell n\left(\dfrac{p+v}{p-v}\right)\\\sf f(p,v)=\dfrac{1}{2}[\ell n (p+v)-\ell n(p-v)]$

$\sf \dfrac{\partial\!~f}{\partial\!~p}=\dfrac{1}{2}\left[\dfrac{1}{p+v}\cdot1-\dfrac{1}{p-v}\cdot1\right]\\\sf\dfrac{\partial\!~f}{\partial\!~p}=\dfrac{1}{2}\left[\dfrac{p-v-1\cdot(p+v)}{(p+v)(p-v)}\right]\\\sf \dfrac{\partial\!~f}{\partial\!~p}=\dfrac{1}{2}\left[\dfrac{\diagup\!\!\!\!p-v-\diagup\!\!\!\!p-v}{p^2-v^2}\right]\\\sf\dfrac{\partial\!~f}{\partial\!~p}=\dfrac{1}{\diagup\!\!\!2}\left[\dfrac{-\diagup\!\!\!2v}{p^2-v^2}\right]$

$\sf\dfrac{\partial\!~f}{\partial\!~p}=\dfrac{-\diagup\!\!\!1\cdot v}{-\diagup\!\!\!1\cdot(-p^2+v^2)}$

$\huge\boxed{\boxed{\boxed{\boxed{\sf\dfrac{\partial\!~f}{\partial\!~p}=\dfrac{v}{v^2-p^2}}}}}$

$\sf\dfrac{\partial\!~f}{\partial\!~v}=\dfrac{1}{2}\cdot\left[\dfrac{1}{p+v}\cdot1-\dfrac{1}{p-v}\cdot-1\right]\\\sf\dfrac{\partial\!~f}{\partial\!~v}=\dfrac{1}{2}\cdot\left[\dfrac{1}{p+v}+\dfrac{1}{p-v}\right]\\\sf\dfrac{\partial\!~f}{\partial\!~v}=\dfrac{1}{2}\cdot\left[\dfrac{p-\diagdown\!\!\!v+p+\diagdown\!\!\!v}{(p+v)(p-v)}\right]\\\sf\dfrac{\partial\!~f}{\partial~v}=\dfrac{1}{\diagup\!\!\!2}\cdot\left[\dfrac{\diagup\!\!\!2p}{p^2-v^2}\right]$

$\huge\boxed{\boxed{\boxed{\boxed{\sf\dfrac{\partial\!~f}{\partial\!~v}=\dfrac{p}{p^2-v^2}\checkmark}}}}$