Question

assume that (f(x,y,z)=xz^{9}+sin(yz^{9}) - 10 = 0) implicitly defines (z) as a function of (x) and (y). find the following partial derivatives: (\frac{partial z}{partial x}=) (\frac{partial z}{partial y}=)

Explanation:

Step1: Recall implicit - differentiation formula

For an implicitly defined function $F(x,y,z) = 0$, $\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}$ 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 $F(x,y,z)=xz^{9}+\sin(y z^{7}) - 10$. Differentiate with respect to $x$ treating $y$ and $z$ as constants. $F_x = z^{9}$.

Step3: Calculate $F_y$

Differentiate $F(x,y,z)$ with respect to $y$ treating $x$ and $z$ as constants. Using the chain - rule, $F_y=z^{7}\cos(y z^{7})$.

Step4: Calculate $F_z$

Differentiate $F(x,y,z)$ with respect to $z$ treating $x$ and $y$ as constants. $F_z = 9xz^{8}+7yz^{6}\cos(y z^{7})$.

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

$\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}=-\frac{z^{9}}{9xz^{8}+7yz^{6}\cos(y z^{7})}=-\frac{z^{3}}{9xz + 7y\cos(y z^{7})}$.

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

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

Answer:

$\frac{\partial z}{\partial x}=-\frac{z^{3}}{9xz + 7y\cos(y z^{7})}$, $\frac{\partial z}{\partial y}=-\frac{z\cos(y z^{7})}{9xz + 7y\cos(y z^{7})}$