Explain why the function is differentiable at the given point. Then find the linearization L(x, y) of the function at that point.

Explain Why The Function Is Differentiable At The Given Point Then Find The Linearization

                                        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:

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

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