Applied Calculus: UNIT II: Functions of Several Variables
Partial derivatives: Problems under type-3
Explanation, Formula, Equation, Example and Solved Problems - Partial derivatives: Problems under type-3
PROBLEMS UNDER TYPE 3
If f(x, y) = 0 is given, dy/dx
= ‒fx/fy where fx = ∂f/∂x,
fy = ∂f/∂y.
Example
29. If 2x2+6xy +4y2+6x+8y + 9 = 0, then find dy/dx
using partial derivatives.
Solution:
Given f(x, y) = 2x2 + 6xy +
4y2 + 6x + 8y + 9
fx
= 4x+6y+6; fy=6x+8y+8
dy/dx = ‒fx/fy = ‒(4x+6y+6) / 6x+8y+8
= ‒2(2x+3y+3) / 2(3x+4y+4)
dy/dx = ‒(2x + 3y+3) / 3x + 4y +4
Example
30. Find dy/dx using partial derivatives if x3 + y3 = 3ax2y.
Solution:
Given f(x, y) = x3+ y3‒3ax2y
fx =
3x2‒6axy;
fy
= 3y2‒3ax2
Example
31. Find dy/dx if x3 + y2 = 3axy.
Solution:
Given f(x, y) = x3 + y3
‒ 3axy
fx
= 3x2 ‒ 3ay; fy
= 3y2 ‒ 3ax
Example
32. If xy + yx = c, then find dy/dx.
Solution:
Given xy +
yx = c.
Let f(x, y) = xy + yx ‒
c
fx
= yxy‒1 + yxlogy;
fy
= xylogx + xyx‒1
Example
33. If x sin(x − y) ‒ (x + y) = 0,
use partial differentiation to prove dy/dx = 
Solution:
Given f(x, y) = x sin(x − y) ‒ (x + y)
∂f/∂x = xcos(x − y) + sin(x − y) – 1
∂f/∂y = xcos(x − y)(− 1) − 1
Applied Calculus: UNIT II: Functions of Several Variables : Tag: Applied Calculus : - Partial derivatives: Problems under type-3
Applied Calculus: UNIT II: Functions of Several Variables
Under Subject
Applied Calculus
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