$$ \mid \leftarrow \quad $$ Find the first partial derivatives of $$ f(x, y)=7 x^{4} y^{6}+9 x^{3} y^{2}+6 $$ $$ \frac{\partial f}{\partial x}=\square $$

Question

UpStudy AI · Accepted Answer

Below we find the first partial derivative of 
$$
f(x, y)=7 x^{4} y^{6}+9 x^{3} y^{2}+6
$$
with respect to $$ x $$.

1. **Differentiate term by term:**

- For the term $$ 7x^4y^6 $$:

The derivative with respect to $$ x $$ is:
     $$
     \frac{\partial}{\partial x} (7x^4y^6) = 7y^6 \cdot \frac{\partial}{\partial x} (x^4) = 7y^6 \cdot 4x^3 = 28x^3y^6.
     $$

- For the term $$ 9x^3y^2 $$:

The derivative with respect to $$ x $$ is:
     $$
     \frac{\partial}{\partial x} (9x^3y^2) = 9y^2 \cdot \frac{\partial}{\partial x} (x^3) = 9y^2 \cdot 3x^2 = 27x^2y^2.
     $$

- The constant term $$ 6 $$ differentiates to $$ 0 $$.

2. **Combine the results:**

$$
   \frac{\partial f}{\partial x} = 28x^3y^6 + 27x^2y^2.
   $$

Thus, the final answer is:

$$
\boxed{28x^3y^6 + 27x^2y^2}
$$
$$ \frac{\partial f}{\partial x} = 28x^3y^6 + 27x^2y^2 $$

Find the first partial derivatives of

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Calculus Questions