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+xIn(xy−5),(2,3)

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Adnaan Franks · Accepted Answer

Step 1 The Linearization L(x,y) for f(x,y) at (x0,y0) is L(x,y)=f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0) Step 2 f(x,y)=1+xln⁡(xy−5) f(2,3)=1+2ln⁡(6−5)=1+2ln⁡1=1 fx=ln⁡(xy−5)+xyxy−5 fx(2,3)=ln⁡(6−5)+66−5=6 fy=x2xy−5 fy(2,3)=226−5=4 Step 3 Linearization at (2,3) is L(x,y)=f(2,3)+fx(2,3)(x−2)+fy(2,3)(y−3)Substitute f(2,3)=1,fx(2,3)=6,fy(2,3)=4 L(x,y)=1+6(x−2)+4(y−3) L(x,y)=1+6x−12+4y−12 L(x,y)=6x+4y−23 Result L(x,y)=6x+4y−23

madeleinejames20 · Accepted Answer

Step 1: Let&#039;s find the partial derivatives of f with respect to x and y:∂f∂x=ln(xy−5)∂f∂y=xxy−5Step 2: Next, let&#039;s evaluate these partial derivatives at the given point (2,3):∂f∂x|(2,3)=ln(2·3−5)=ln(1)=0∂f∂y|(2,3)=22·3−5=21=2Since both partial derivatives exist and are continuous at (2,3), the function f(x,y) is differentiable at that point.Step 3: Now, let&#039;s find the linearization L(x,y) of the function f at the point (2,3). The linearization is given by:L(x,y)=f(2,3)+∂f∂x|(2,3)(x−2)+∂f∂y|(2,3)(y−3)Substituting the values we found earlier:L(x,y)=1+0(x−2)+2(y−3)=1+2(y−3)Therefore, the linearization of the function f(x,y) at the point (2,3) is L(x,y)=1+2(y−3).

Eliza Beth13 · Accepted Answer

Answer:4x+4y−19Explanation:To show why the function is differentiable at the given point (2,3) and find its linearization L(x,y), we need to evaluate the partial derivatives and substitute the point&#039;s coordinates into the linearization formula.Let&#039;s start by calculating the partial derivatives of f(x,y) with respect to x and y using the product rule and chain rule:∂f∂x=∂∂x(1+xln(xy−5))=ln(xy−5)+x·∂∂xln(xy−5)=ln(xy−5)+x·1xy−5·∂∂x(xy−5)=ln(xy−5)+x·1xy−5·(y)=ln(xy−5)+xyxy−5Now, let&#039;s find the partial derivative with respect to y:∂f∂y=∂∂y(1+xln(xy−5))=x·∂∂yln(xy−5)=x·1xy−5·∂∂y(xy−5)=x·1xy−5·(x)=x2xy−5Now, let&#039;s substitute the coordinates (2,3) into the partial derivatives:∂f∂x(2,3)=ln(2·3−5)+2·22·3−5=ln(6−5)+46−5=ln(1)+4=0+4=4∂f∂y(2,3)=222·3−5=46−5=41=4Now, we can write the linearization L(x,y) using the point-slope form of a linear equation:L(x,y)=f(2,3)+∂f∂x(2,3)·(x−2)+∂f∂y(2,3)·(y−3)L(x,y)=f(2,3)+4·(x−2)+4·(y−3)L(x,y)=f(2,3)+4x−8+4y−12L(x,y)=f(2,3)+4x+4y−20Since we&#039;re given the point (2,3), we substitute these values into the original function to find f(2,3):f(2,3)=1+2ln(2·3−5)=1+2ln(6−5)=1+2ln(1)=1+2·0=1+0=1Therefore, the linearization L(x,y) of the function at the point (2,3) is:L(x,y)=1+4x+4y−20L(x,y)=4x+4y−19.

Nick Camelot · Accepted Answer

To determine the differentiability of the function f(x,y)=1+xln(xy−5) at the point (2,3), we need to check if the partial derivatives of f exist and are continuous in a neighborhood of that point.First, let&#039;s find the partial derivatives of f with respect to x and y:∂f∂x=∂∂x(1+xln(xy−5))Using the product rule, we have:∂f∂x=0+ln(xy−5)+x·∂∂x(ln(xy−5))Now, let&#039;s find ∂∂x(ln(xy−5)):∂∂x(ln(xy−5))=∂∂x(ln(xy−5))·∂(xy−5)∂xSimplifying further:∂∂x(ln(xy−5))=∂∂x(ln(xy−5))·yTherefore, we have:∂f∂x=ln(xy−5)+x·∂∂x(ln(xy−5))=ln(xy−5)+x·∂∂x(ln(xy−5))·yNext, let&#039;s find ∂f∂y:∂f∂y=∂∂y(1+xln(xy−5))Applying the chain rule:∂f∂y=x·∂∂y(ln(xy−5))=x·∂∂y(ln(xy−5))·∂(xy−5)∂ySimplifying further:∂f∂y=x·∂∂y(ln(xy−5))·x=x·∂∂y(ln(xy−5))·xNow, let&#039;s evaluate the partial derivatives at the point (2,3):∂f∂x|(2,3)=ln(2·3−5)+2·∂∂x(ln(2·3−5))·3∂f∂y|(2,3)=2·∂∂y(ln(2·3−5))·2Now, let&#039;s find the values of the partial derivatives at the point (2,3):∂f∂x|(2,3)=ln(6−5)+2·∂∂x(ln(6−5))·3=ln(1)+2·16−5·3=ln(1)+2·31=0+2·3=6∂f∂y|(2,3)=2·∂∂y(ln(6−5))·2=2·16−5·2=2·2=4Since both partial derivatives exist and are continuous at the point (2,3), the function f is differentiable at that point.Next, let&#039;s find the linearization L(x,y) of the function f(x,y) at the point (2,3). The linearization can be expressed using the equation of a plane:L(x,y)=f(2,3)+∂f∂x|(2,3)·(x−2)+∂f∂y|(2,3)·(y−3)Substituting the values we found earlier:L(x,y)=f(2,3)+6·(x−2)+4·(y−3)Therefore, the linearization of the function f(x,y) at the point (2,3) is:L(x,y)=f(2,3)+6·(x−2)+4·(y−3)

Explain why the function is differentiable at the given point. Then find the lin

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