The problem asks for the value of the partial derivative $\frac{\partial h(x,y)}{\partial x}$ at the point $x = 4.3$ and $y = 4.1$ for the function given by:

$h(x, y) = 2xy + 3x^2 + 4x^3y^2 + 8$

Step 1: Compute the partial derivative of $h(x, y)$ with respect to $x$

We need to differentiate $h(x, y)$ with respect to $x$ while treating $y$ as a constant.

$\frac{\partial h(x,y)}{\partial x} = \frac{\partial}{\partial x} \left(2xy + 3x^2 + 4x^3y^2 + 8\right)$

Differentiate each term:

• The derivative of $2xy$ with respect to $x$ is $2y$.
• The derivative of $3x^2$ with respect to $x$ is $6x$.
• The derivative of $4x^3y^2$ with respect to $x$ is $12x^2y^2$.
• The derivative of the constant $8$ with respect to $x$ is $0$.

So, the partial derivative $\frac{\partial h(x,y)}{\partial x}$ is:

$\frac{\partial h(x,y)}{\partial x} = 2y + 6x + 12x^2y^2$

Step 2: Substitute $x = 4.3$ and $y = 4.1$ into the partial derivative

$\frac{\partial h(4.3,4.1)}{\partial x} = 2(4.1) + 6(4.3) + 12(4.3)^2(4.1)^2$

Calculate each term:

$2(4.1) = 8.2$

$6(4.3) = 25.8$

$12(4.3)^2(4.1)^2 = 12 \times 18.49 \times 16.81 \approx 12 \times 310.757 \approx 3729.084$

Adding them up:

$8.2 + 25.8 + 3729.084 \approx 3763.084$

Final Answer:

The value of $\frac{\partial h(x,y)}{\partial x}$ at $x = 4.3$ and $y = 4.1$ is approximately 3763.084.

Would you like further details or have any questions? Here are some related questions to deepen your understanding:

1. How do you find the partial derivative with respect to $y$ for the same function?
2. What is the significance of the partial derivative in multivariable functions?
3. How does the value of the partial derivative change if we slightly alter the values of $x$ and $y$?
4. Can you compute the second partial derivative $\frac{\partial^2 h}{\partial x^2}$?
5. What happens to the function $h(x, y)$ when $y$ is held constant, and only $x$ changes?

Tip: When dealing with partial derivatives, always remember to treat the other variables as constants. This approach simplifies the process and helps avoid mistakes.