Linear Approximation: Find & Use for f(2.28,8.22)

[andyk23]

Find the linear approximation to the equation f(x,y) = 3 sqrt((x y)/4) at the point (2,8,6), and use it to approximate f(2.28,8.22). I know you take the derivative of fx(x,y) and fy(x,y), I think I'm taking the derivative wrong. Then after that you put x and y in the equation and solve for fx(2,8) and fy(2,8). Then take f(a,b)+fx(a,b)(x-a)+fy(a,b)(y-b). For fx(x,y) I'm coming up with .75(y/4) and for fy(x,y) I'm getting .75(x/4)

 

[jbunniii]

Your derivatives are indeed wrong. Let's start with a simpler function of just one variable.

$$g(x) = 3 \sqrt{x / 4}$$What is $g'(x)$?

 

[andyk23]

g(x)=3(x/4)^-.5
g'(x)= [3(x)^-.5]/4

 

[Rayquesto]

Want to know an easy way to derive a linear approximation equation? Well, I'll give you some intuition: it's basically the same thing as the tangent line equation in so many words.

 

[Rayquesto]

L(x)= f(x) + f'(a)(x-a)

(2.8,6) is the point...and so on