Math Problem Statement

  1. Let f (x, y) = xyxy sin(xy) + x3y+x4 y2+xy+1 arctan( ln(x) x5+y6 ). Use the limit definition of ∂f ∂y to find ∂f ∂y (1, π). Note: Do not try to find closed form formula for ∂

Solution

To find the partial derivative of f(x,y)f(x, y) with respect to yy at the point (1,π)(1, \pi), using the limit definition of the derivative, we'll use:

fy(x0,y0)=limh0f(x0,y0+h)f(x0,y0)h\frac{\partial f}{\partial y} (x_0, y_0) = \lim_{h \to 0} \frac{f(x_0, y_0 + h) - f(x_0, y_0)}{h}

Given:

  • Function: f(x,y)=xysin(xy)+x3y+x4y2+xy+1arctan(ln(x)x5+y6)f(x, y) = xy \sin(xy) + x^3 y + x^4 y^2 + xy + 1 \arctan \left( \frac{\ln(x)}{x^5 + y^6} \right)
  • Point: (x0,y0)=(1,π)(x_0, y_0) = (1, \pi)

We want to find:

fy(1,π)\frac{\partial f}{\partial y} (1, \pi)

Step-by-Step Computation

Step 1: Write the Limit Definition at (1,π)(1, \pi)

fy(1,π)=limh0f(1,π+h)f(1,π)h\frac{\partial f}{\partial y} (1, \pi) = \lim_{h \to 0} \frac{f(1, \pi + h) - f(1, \pi)}{h}

Step 2: Substitute the Function

We will evaluate the function f(1,y)f(1, y) first. This means substituting x=1x = 1 into the function:

f(1,y)=1ysin(1y)+13y+14y2+1y+1arctan(ln(1)15+y6)f(1, y) = 1 \cdot y \sin(1 \cdot y) + 1^3 y + 1^4 y^2 + 1 \cdot y + 1 \arctan\left( \frac{\ln(1)}{1^5 + y^6} \right)

Simplify each term:

  • ysin(y)y \sin(y)
  • +y+ y
  • +y2+ y^2
  • +y+ y
  • Since ln(1)=0\ln(1) = 0, the term involving arctan\arctan is zero.

Thus:

f(1,y)=ysin(y)+y+y2+y=ysin(y)+2y+y2f(1, y) = y \sin(y) + y + y^2 + y = y \sin(y) + 2y + y^2

Step 3: Apply the Limit Definition

We need to compute:

fy(1,π)=limh0f(1,π+h)f(1,π)h\frac{\partial f}{\partial y} (1, \pi) = \lim_{h \to 0} \frac{f(1, \pi + h) - f(1, \pi)}{h}

Substituting the simplified form of f(1,y)f(1, y):

f(1,π+h)=(π+h)sin(π+h)+2(π+h)+(π+h)2f(1, \pi + h) = (\pi + h) \sin(\pi + h) + 2(\pi + h) + (\pi + h)^2

Now expand:

  1. (π+h)sin(π+h)(\pi + h) \sin(\pi + h)
  2. +2(π+h)=2π+2h+ 2(\pi + h) = 2\pi + 2h
  3. +(π+h)2=π2+2πh+h2+ (\pi + h)^2 = \pi^2 + 2\pi h + h^2

Thus:

f(1,π+h)=(π+h)sin(π+h)+2π+2h+π2+2πh+h2f(1, \pi + h) = (\pi + h) \sin(\pi + h) + 2\pi + 2h + \pi^2 + 2\pi h + h^2

Similarly, evaluate f(1,π)f(1, \pi):

f(1,π)=πsin(π)+2π+π2f(1, \pi) = \pi \sin(\pi) + 2\pi + \pi^2

Since sin(π)=0\sin(\pi) = 0:

f(1,π)=2π+π2f(1, \pi) = 2\pi + \pi^2

Step 4: Find the Difference f(1,π+h)f(1,π)f(1, \pi + h) - f(1, \pi)

f(1,π+h)f(1,π)=[(π+h)sin(π+h)+2π+2h+π2+2πh+h2](2π+π2)f(1, \pi + h) - f(1, \pi) = [(\pi + h) \sin(\pi + h) + 2\pi + 2h + \pi^2 + 2\pi h + h^2] - (2\pi + \pi^2)

Simplify:

f(1,π+h)f(1,π)=(π+h)sin(π+h)+2h+2πh+h2f(1, \pi + h) - f(1, \pi) = (\pi + h) \sin(\pi + h) + 2h + 2\pi h + h^2

Step 5: Divide by hh and Take the Limit

f(1,π+h)f(1,π)h=(π+h)sin(π+h)+2h+2πh+h2h\frac{f(1, \pi + h) - f(1, \pi)}{h} = \frac{(\pi + h) \sin(\pi + h) + 2h + 2\pi h + h^2}{h}

Split into individual terms:

=(π+h)sin(π+h)h+2hh+2πhh+h2h= \frac{(\pi + h) \sin(\pi + h)}{h} + \frac{2h}{h} + \frac{2\pi h}{h} + \frac{h^2}{h}

Simplify each term:

=(π+h)sin(π+h)h+2+2π+h= \frac{(\pi + h) \sin(\pi + h)}{h} + 2 + 2\pi + h

Take the limit as h0h \to 0:

  • limh0(π+h)sin(π+h)h=sin(π)+πcos(π)=0π=π\lim_{h \to 0} \frac{(\pi + h) \sin(\pi + h)}{h} = \sin(\pi) + \pi \cos(\pi) = 0 - \pi = -\pi
  • The other terms become 2+2π2 + 2\pi

Thus:

fy(1,π)=π+2+2π=π+2\frac{\partial f}{\partial y} (1, \pi) = -\pi + 2 + 2\pi = \pi + 2

Final Answer

fy(1,π)=π+2\frac{\partial f}{\partial y} (1, \pi) = \pi + 2


Would you like more details or do you have any questions?

Here are some related questions to expand your understanding:

  1. How is the limit definition of a partial derivative different from the regular derivative?
  2. What role does the sine function play in determining the complexity of the limit evaluation?
  3. How can we generalize this method to calculate higher-order partial derivatives?
  4. Can you explain why we set x=1x = 1 early in the process instead of differentiating with respect to yy first?
  5. How do trigonometric properties affect the computation of partial derivatives in general?

Tip: When applying the limit definition for partial derivatives, make sure to carefully simplify the expressions to avoid algebraic errors, especially when trigonometric or logarithmic functions are involved.

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Math Problem Analysis

Mathematical Concepts

Partial Derivatives
Multivariable Calculus
Trigonometric Functions
Limit Definition

Formulas

∂f/∂y(x_0, y_0) = lim(h→0) [(f(x_0, y_0 + h) - f(x_0, y_0)) / h]
f(x, y) = xy sin(xy) + x^3y + x^4y^2 + xy + arctan(ln(x) / (x^5 + y^6))

Theorems

Limit Definition of Partial Derivatives
Trigonometric Limits

Suitable Grade Level

College-Level Calculus