Question

Calcular derivada de z em relação a v, conforme exercício na imagem anexa.

Answer 1

$\boxed{\dfrac{dz}{dv}=\dfrac{dz}{dy}\cdot \dfrac{dy}{dx}\cdot \dfrac{dx}{dw}\cdot \dfrac{dw}{dv}}$


Para facilitar, vou começar da derivada da última função para a da primeira:

$\bullet\;\;\dfrac{dw}{dv}=\dfrac{d}{dv}\left[\mathrm{\ell n}\left(3^{4v+7} \right ) \right ]\\ \\ \dfrac{dw}{dv}=\dfrac{1}{3^{4v+7}}\cdot \dfrac{d}{dv}\left(3^{4v+7} \right )\\ \\ \dfrac{dw}{dv}=\dfrac{1}{3^{4v+7}}\cdot \left(3^{4v+7}\cdot \mathrm{\ell n\,}3\right)\cdot \dfrac{d}{dv}\left(4v+7 \right )\\ \\ \dfrac{dw}{dv}=\mathrm{\ell n\,}3\cdot \left(4+0\right)\\ \\ \boxed{\dfrac{dw}{dv}=4\mathrm{\,\ell n\,}3}$


$\bullet\;\;\dfrac{dx}{dw}=\dfrac{d}{dw}\left(\sec \left(w \right )-1 \right )\\ \\ \dfrac{dx}{dw}=\sec\left(w \right )\cdot\mathrm{tg}\left(w \right )+0\\ \\ \boxed{ \dfrac{dx}{dw}=\sec\left[\mathrm{\ell n}\left(3^{4v+7} \right ) \right ]\cdot\mathrm{tg}\left[\mathrm{\ell n}\left(3^{4v+7} \right ) \right ]}$


$\bullet\;\;\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\sqrt{\left(x+1 \right )^{3}} \right )\\ \\ \dfrac{dy}{dx}=\dfrac{d}{dx}\left(\left(x+1 \right )^{\,^{3}\!\!\!\diagup\!\!_{2}} \right )\\ \\ \dfrac{dy}{dx}=\dfrac{3}{2}\cdot \left(x+1 \right )^{\,^{3}\!\!\!\diagup\!\!_{2}-1}\dfrac{d}{dx}\left(x+1 \right )\\ \\ \dfrac{dy}{dx}=\dfrac{3}{2}\cdot \left(x+1 \right )^{\,^{1}\!\!\!\diagup\!\!_{2}}\cdot 1\\ \\ \dfrac{dy}{dx}=\dfrac{3}{2}\cdot \left[\sec\left(w \right )-\diagup\!\!\!\!1+\diagup\!\!\!\!1 \right ]^{\,^{1}\!\!\!\diagup\!\!_{2}}\\ \\ \dfrac{dy}{dx}=\dfrac{3}{2}\cdot \left[\sec\left(\mathrm{\ell n}\left(3^{4v+7} \right ) \right ) \right ]^{\,^{1}\!\!\!\diagup\!\!_{2}}\\ \\ \boxed{\dfrac{dy}{dx}=\dfrac{3}{2}\cdot \sqrt{\sec\left[\mathrm{\ell n}\left(3^{4v+7} \right ) \right ]}}$


$\bullet\;\;\dfrac{dz}{dy}=\dfrac{d}{dy}\left(y^{2}+6y-1 \right )\\ \\ \dfrac{dz}{dy}=2y+6+0\\ \\ \dfrac{dz}{dy}=2\sqrt{\left(x+1 \right )^{3}}+6\\ \\ \dfrac{dz}{dy}=2\sqrt{\left(\sec \left(w \right )-\diagup\!\!\!\!1+\diagup\!\!\!\!1 \right )^{3}}+6\\ \\ \dfrac{dz}{dy}=2\sqrt{\sec^{3} \left(w \right )}+6\\ \\ \boxed{\dfrac{dz}{dy}=2\sqrt{\sec^{3} \left[\mathrm{\ell n}\left(3^{4v+7} \right ) \right ]}+6}$


Então, finalmente temos

$\dfrac{dz}{dv}=\dfrac{dz}{dy}\cdot \dfrac{dy}{dx}\cdot \dfrac{dx}{dw}\cdot \dfrac{dw}{dv}\\ \\ \boxed{\begin{array}{rcl} \dfrac{dz}{dv}&=&\left(2\sqrt{\sec^{3} \left[\mathrm{\ell n}\left(3^{4v+7} \right ) \right ]}+6 \right )\cdot \left(\dfrac{3}{2}\cdot \sqrt{\sec\left[\mathrm{\ell n}\left(3^{4v+7} \right ) \right ]} \right ) \cdot\\ \\ &&\cdot \left(\sec\left[\mathrm{\ell n}\left(3^{4v+7} \right ) \right ]\mathrm{tg}\left[\mathrm{\ell n}\left(3^{4v+7} \right ) \right ] \right )\cdot \left(4\mathrm{\,\ell n\,}3 \right ) \end{array}}$