Partial Derivative Calculation

This document demonstrates the calculation of the partial derivative of a function with respect to x.

The step-by-step solution shows the application of differentiation rules in multivariable calculus.

Original Function

\[f(x, y) = -(x-3)^2 + xy + 16\]

This is a quadratic function in x with a linear term in y. We will find the partial derivative with respect to x.

Partial Derivative with Respect to x

1 \[\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left[ -(x-3)^2 + xy + 16 \right]\]

Start with the definition of the partial derivative with respect to x.

2 \[= \frac{\partial}{\partial x} \left[ -(x-3)^2 \right] + \frac{\partial}{\partial x} \left[ xy \right] + \frac{\partial}{\partial x} \left[ 16 \right]\]

Apply the sum rule: the derivative of a sum is the sum of derivatives.

3 \[= -2(x-3) \frac{\partial}{\partial x} (x-3) + y \frac{\partial}{\partial x} (x) + 0\]

- Apply the chain rule to differentiate \(-(x-3)^2\)
- Apply the constant multiple rule to differentiate \(xy\) (treating y as constant)
- The derivative of a constant (16) is zero

4 \[= -2(x-3)(1) + y(1)\]

- The derivative of \((x-3)\) with respect to x is 1
- The derivative of \(x\) with respect to x is 1

5 \[= -2(x-3) + y\]

Simplify the expression by multiplying.

6 \[= -2x - 2(-3) + y\]

Distribute the -2 through the parentheses.

7 \[= -2x + y + 6\]

Simplify the constant term to get the final result.

Final Result

\[\frac{\partial f}{\partial x} = -2x + y + 6\]

The partial derivative of f(x,y) with respect to x is a linear function in both x and y.