** f(x,y) = 1 + x ln (xy – 5), (2,3)**

This problem explains why the given function is **differentiable** at a **point,** and to find the **linearization** at that **point.** The concept required to solve this problem includes the **method** for finding **partial derivatives** **fx** and **fy** of the function **z = f(x,y)**, the **partial derivatives theorem**, and the equation of **linearization.**

The** theorem of partial derivatives** states that if the **partial derivatives** **fx** and **fy** are **continuous** and exist **near** a point **(a,b)**, the function is **differentiable** at that point.

**Linearization** is the method of finding the **linear approximation** of a function $f(x,y)$ at a given point $(a,b)$ with the **formula:**

\[ L(x,y)=f(a,b)+(x-a)f_x(a,b)+(y-b)f_y(a,b)\]

The above equation is similar to the **one variable linear** equation $L(x)=f(a)+f'(a)(x-a)$.

## Expert Answer

Given the **equation:**

\[ f(x,y) = 1 + x \ln (xy-5); \space \text{and the point is}\space (2,3)\]

Therefore,

\[ f(2,3) = 1 + 2 \ln ((2)(3)-5) \]

\[ f(2,3) = 1 \]

First, we will find the **partial derivatives** of $f$ in order to use the **theorem.**

**Differentiating** the equation $ f(x,y) = 1 + x \ln (xy-5)$ with **respect** to $x$ to find $f_x$:

\[ f(x,y) = 1 + x \ln (xy-5)\]

\[ f_x(x,y) = x \times \dfrac{1}{xy-5}(y) + \ln (xy-5) \times 1 \]

That is,

\[ f_x(x,y) = \dfrac{xy}{xy-5} + \ln (xy-5) \]

**Putting** $(2,3)$:

\[ f_x(2,3) = \dfrac{(2)(3)}{(2)(3)-5} + \ln ((2)(3)-5) \]

\[ f_x(x,y) = 6 +\ln (1) \]

\[ f_x(x,y) = 6 \]

Now **differentiate** with **respect** to $y$ to find $f_y$:

\[ f_y(x,y) = x \times \dfrac{1}{xy-5}(x) \]

Becomes,

\[ f_y(x,y) = \dfrac{x^2}{xy-5} \]

**Putting** $(2,3)$:

\[ f_y(x,y) = \dfrac{2^2}{(2)(3)-5} \]

\[ f_y(x,y) = 4 \]

Hence, we **conclude** that $f_x(x,y) = \dfrac{xy}{xy-5} + \ln (xy-5)$ and $f_y(x,y) = \dfrac{x^2}{xy-5}$ **exist,** and are **continuous** for $x\geq 5$, which **means** both $f_x$ and $f_y$ are **continuous** and **exist** near the **point** $(2,3)$.

Therefore,

\[ f(x,y) = 1 + x \ln (xy-5); \space \text{is differentiable at point} \space (2,3)\]

Now, using the **linearization equation:**

\[ L(x,y) = f(2,3) + (x-2)f_x(2,3) + (y-3)f_y(2,3) \]

**Substituting** the values:

\[ L(x,y) = 1 + (x-2)(6) + (y-3)(4) \]

Hence, the **linearization function** is:

\[ L(x,y) = 6x + 4y – 23 \]

## Numerical Result

$f(x,y)$ is **differentiable** at the **point** $(2,3)$ and the **linearization** of $f(2,3)$ is $L(x,y) = 6x + 4y – 23$.

## Example

Give a reason for the **function** to be **differentiable** at the given **point,** and also find the **linearization** of the **function** at the same point.

$f(x,y)=\dfrac{1+y}{1+x};\space (1,3)$

Rearrange the **function:**

\[ f(x,y) = (1+y)(1+x)^{-1}\]

The **partial derivates** are:

\[ f_x(x,y) = (1+y)(-1)(1+x)^{-2}\]

\[ f_x(x,y) = – \dfrac{1+y}{(1+x)^2}\]

And,

\[f_y(x,y) = (1)(1+x)^{-1}\]

\[f_y(x,y) = – \dfrac{1}{1+x}\]

Now, **substituting** the **point:**

\[f_x(1,3) = – \dfrac{1+3}{(1+1)^2}\]

\[f_x(1,3) = – 1\]

Similarly,

\[f_y(1,3) = – \dfrac{1}{1+1}\]

\[f_x(1,3)=\dfrac{1}{2}\]

Both $f_x$ and $f_y$ are **continuous functions** for $x \neq -1$, so $f$ is **differentiable** at point $(1,3)$.

Now, using the **linearization equation:**

\[L(x,y)=f(1,3) + (x-1)f_x(1,3) + (y-3)f_y(1,3) \]

**Substituting** the values:

\[L(x,y)=2 + (x-1)(-1) + (y-3)(\dfrac{1}{2}) \]

Hence, the **linearization function** is:

\[L(x,y)=-x + \dfrac{1}{2}y + \dfrac{3}{2}\]