# Math Help - An Approximation to a partial Derivative

1. ## An Approximation to a partial Derivative

An Approximation to a partial Derivative

If a function is known to have
fx(30,24) = -.4
fy(30,24) = 1.2
f(30,24) = 50

Estimate this value of F.
f(30.2,23.9)

Thank you for helping me. I am stuck on the last part

2. Originally Posted by Applestar13
An Approximation to a partial Derivative

If a function is known to have
fx(30,24) = -.4
fy(30,24) = 1.2
f(30,24) = 50

Estimate this value of F.
f(30.2,23.9)

Thank you in advance! I am stuck on this part. Both s and y changed and have no idea what to do.
the idea of approximating a two variable function around a given point is to use the tangent plane (at the given point) to the surface that the function represents as

the approximation, i.e. $f(x,y) \approx f(x_0,y_0)+(x-x_0)f_x(x_0,y_0)+(y-y_0)f_y(x_0,y_0).$ so if we put $x=30.2, \ y=23.9, \ x_0=30, \ y_0=24,$ and use the given info, we'll

get: $f(30.2,23.9) \approx 50 + (0.2) \times (-0.4) + (-0.1) \times (1.2) = 49.8.$

3. Originally Posted by Applestar13
An Approximation to a partial Derivative

If a function is known to have
fx(30,24) = -.4
fy(30,24) = 1.2
f(30,24) = 50

Estimate this value of F.
f(30.2,23.9)

Thank you for helping me. I am stuck on the last part
$
\nabla f(30,24) = \left[ - 0.4,1.2 \right]$

$
f(30.2,23.9) \approx f(30,24) + [0.2, - 0.1].\nabla f(30,24)$

.................. $
= 50 + ( - 0.2 \times 0.4 - 0.1 \times 1.2) = 49.8
$

RonL

4. ## Thank you

Thank you, both of you. That made perfect sense. I didn't know what to do with the difference in the intial value of X and Y.