Question

assume that $f(x,y,z)=x^{9}z^{7}+sin(y^{9}z^{7}) - 10 = 0$ implies that $z$ is defined as a function of $x$ and $y$. find the following partial derivatives: $\frac{partial z}{partial x}=\frac{9x^{8}z^{7}}{-(x^{9}+y^{9}cos(y^{9}z^{7}))}$, $\frac{partial z}{partial y}=\frac{-9y^{8}zcos(y^{9}z^{7})}{(x^{9}+y^{9}cos(y^{9}z^{7}))}$

Explanation:

Step1: Recall implicit - differentiation formula

If $F(x,y,z) = 0$, 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 $F(x,y,z)=x^{9}z^{7}+\sin(y^{9}z^{7}) - 10$. Differentiate with respect to $x$ treating $y$ and $z$ as constants. Using the power - rule, $F_x = 9x^{8}z^{7}$.

Step3: Calculate $F_y$

Differentiate $F(x,y,z)$ with respect to $y$ treating $x$ and $z$ as constants. Using the chain - rule, if $u = y^{9}z^{7}$, then $\frac{\partial}{\partial y}\sin(u)=\cos(u)\cdot\frac{\partial u}{\partial y}$. So $F_y=9y^{8}z^{7}\cos(y^{9}z^{7})$.

Step4: Calculate $F_z$

Differentiate $F(x,y,z)$ with respect to $z$ treating $x$ and $y$ as constants. Using the product - rule for $x^{9}z^{7}$ (where $(uv)^\prime = u^\prime v+uv^\prime$ with $u = x^{9}$ and $v = z^{7}$) and the chain - rule for $\sin(y^{9}z^{7})$. $F_z=7x^{9}z^{6}+7y^{9}z^{6}\cos(y^{9}z^{7})$.

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

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

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

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

Answer:

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