Question

assume that $f(x,y,z)=x^{9}z^{7}+sin(y^{9}z^{7}) - 10 = 0$. implicitly defines $z$ as a function of $x$ and $y$. find the following partial derivatives: $\frac{partial z}{partial x}=-\frac{f_{x}}{f_{z}}$ and $\frac{partial z}{partial y}=-\frac{f_{y}}{f_{z}}$. the multivariable chain rule can be used to show that $\frac{partial z}{partial x}=-\frac{9xz + 7ycos(yz^{7})}{z^{3}}$ $\frac{partial z}{partial y}=-\frac{zcos(yz^{7})}{9xz + 7ycos(yz^{7})}$ hide hint

Explanation:

Step1: Recall the formula for implicit - differentiation

If $F(x,y,z) = 0$ implicitly defines $z$ as a function of $x$ and $y$, then $\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}$ and $\frac{\partial z}{\partial y}=-\frac{F_y}{F_z}$, where $F_x=\frac{\partial F}{\partial x}$, $F_y=\frac{\partial F}{\partial y}$, and $F_z=\frac{\partial F}{\partial z}$.

Step2: Calculate $F_x$

Given $F(x,y,z)=x^9z^7+\sin(y^9z^7)-10$. Using the product - rule and the chain - rule, $F_x = 9x^8z^7$.

Step3: Calculate $F_y$

$F_y=\cos(y^9z^7)\cdot9y^8z^7$.

Step4: Calculate $F_z$

$F_z = 7x^9z^6+7y^9z^6\cos(y^9z^7)=7z^6(x^9 + y^9\cos(y^9z^7))$.

Step5: Calculate $\frac{\partial z}{\partial x}$

$\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}=-\frac{9x^8z^7}{7z^6(x^9 + y^9\cos(y^9z^7))}=\frac{- 9x^8z}{7(x^9 + y^9\cos(y^9z^7))}$.

Step6: Calculate $\frac{\partial z}{\partial y}$

$\frac{\partial z}{\partial y}=-\frac{F_y}{F_z}=-\frac{9y^8z^7\cos(y^9z^7)}{7z^6(x^9 + y^9\cos(y^9z^7))}=\frac{-9y^8z\cos(y^9z^7)}{7(x^9 + y^9\cos(y^9z^7))}$.

Answer:

$\frac{\partial z}{\partial x}=\frac{-9x^8z}{7(x^9 + y^9\cos(y^9z^7))}$, $\frac{\partial z}{\partial y}=\frac{-9y^8z\cos(y^9z^7)}{7(x^9 + y^9\cos(y^9z^7))}$