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

                                        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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