Question

find the first partial derivatives of the function $z=(2x + 6y)^{10}$. 1. $\frac{partial z}{partial x}=$ 2. $\frac{partial z}{partial y}=$

Explanation:

Step1: Recall chain - rule for partial derivatives

If $z=(2x + 6y)^{10}$, let $u = 2x+6y$, then $z = u^{10}$. The chain - rule for partial derivatives $\frac{\partial z}{\partial x}=\frac{dz}{du}\cdot\frac{\partial u}{\partial x}$. First, find $\frac{dz}{du}$ and $\frac{\partial u}{\partial x}$.
$\frac{dz}{du}=10u^{9}$ and $\frac{\partial u}{\partial x}=2$.

Step2: Substitute $u = 2x + 6y$ back

$\frac{\partial z}{\partial x}=10(2x + 6y)^{9}\cdot2=20(2x + 6y)^{9}$.

Step3: For $\frac{\partial z}{\partial y}$ using chain - rule

Again, with $u = 2x+6y$ and $z = u^{10}$, $\frac{\partial z}{\partial y}=\frac{dz}{du}\cdot\frac{\partial u}{\partial y}$. We know that $\frac{dz}{du}=10u^{9}$ and $\frac{\partial u}{\partial y}=6$.

Step4: Substitute $u$ back

$\frac{\partial z}{\partial y}=10(2x + 6y)^{9}\cdot6=60(2x + 6y)^{9}$.

Answer:

1. $\frac{\partial z}{\partial x}=20(2x + 6y)^{9}$
2. $\frac{\partial z}{\partial y}=60(2x + 6y)^{9}$