find $\frac{partial z}{partial y}$ if $z=(7x + 9y)^3$.\n$\frac{partial z}{partial y}=square$

Explanation:

Step1: Apply chain - rule

Let $u = 7x+9y$, then $z = u^{3}$. The chain - rule for partial derivatives is $\frac{\partial z}{\partial y}=\frac{dz}{du}\cdot\frac{\partial u}{\partial y}$. First, find $\frac{dz}{du}$. Since $z = u^{3}$, by the power rule $\frac{dz}{du}=3u^{2}$.

Step2: Find $\frac{\partial u}{\partial y}$

Since $u = 7x + 9y$, and treating $x$ as a constant, $\frac{\partial u}{\partial y}=9$.

Step3: Calculate $\frac{\partial z}{\partial y}$

Substitute $\frac{dz}{du}$ and $\frac{\partial u}{\partial y}$ into the chain - rule formula: $\frac{\partial z}{\partial y}=\frac{dz}{du}\cdot\frac{\partial u}{\partial y}=3u^{2}\cdot9$. Replace $u = 7x + 9y$ back in, we get $\frac{\partial z}{\partial y}=27(7x + 9y)^{2}$.

Answer:

$27(7x + 9y)^{2}$