Math Problem Statement

The equation 4(xy^2+y^2)^(1/4)+xy−20=0

defines the variable y as a function of variable x (y=f(x) ) in the neighborhood of the point (0,5) . Evaluate y′(0) . remember to use the product rule and the chain rule

Solution

To find y(0)y'(0), we need to implicitly differentiate the given equation with respect to xx and then solve for y(x)y'(x) at x=0x = 0 and y=5y = 5.

The equation is: 4(xy2+y2)14+xy20=04(xy^2 + y^2)^{\frac{1}{4}} + xy - 20 = 0

  1. Differentiate each term with respect to xx:

    Using the product rule, chain rule, and implicit differentiation where y=y(x)y = y(x):

    • First term: 4(xy2+y2)144(xy^2 + y^2)^{\frac{1}{4}} ddx(4(xy2+y2)14)=414(xy2+y2)34ddx(xy2+y2)\frac{d}{dx} \left(4(xy^2 + y^2)^{\frac{1}{4}}\right) = 4 \cdot \frac{1}{4} (xy^2 + y^2)^{-\frac{3}{4}} \cdot \frac{d}{dx}(xy^2 + y^2) =(xy2+y2)34(y2+x2yy+2yy)= (xy^2 + y^2)^{-\frac{3}{4}} \cdot \left( y^2 + x \cdot 2y \cdot y' + 2y \cdot y' \right) Simplifying, =(xy2+y2)34(y2+2xyy+2yy)= (xy^2 + y^2)^{-\frac{3}{4}} \cdot (y^2 + 2xyy' + 2yy')

    • Second term: xyxy ddx(xy)=y+xy\frac{d}{dx}(xy) = y + x \cdot y'

    • Third term: Constant term 20-20 ddx(20)=0\frac{d}{dx}(-20) = 0

  2. Combine derivatives:

    Substitute the differentiated terms into the equation: (xy2+y2)34(y2+2xyy+2yy)+y+xy=0(xy^2 + y^2)^{-\frac{3}{4}} \cdot (y^2 + 2xyy' + 2yy') + y + xy' = 0

  3. Evaluate at x=0x = 0 and y=5y = 5:

    Substitute x=0x = 0, y=5y = 5, and y(0)=yy'(0) = y' into the equation: (052+52)34(52+205y+25y)+5+0y=0(0 \cdot 5^2 + 5^2)^{-\frac{3}{4}} \cdot (5^2 + 2 \cdot 0 \cdot 5 \cdot y' + 2 \cdot 5 \cdot y') + 5 + 0 \cdot y' = 0 Simplifying, (25)34(25+10y)+5=0(25)^{-\frac{3}{4}} \cdot (25 + 10y') + 5 = 0

  4. Solve for yy':

    1253/4(25+10y)+5=0\frac{1}{25^{3/4}} (25 + 10y') + 5 = 0 Since 253/4=53=12525^{3/4} = 5^{3} = 125, we have: 1125(25+10y)+5=0\frac{1}{125} (25 + 10y') + 5 = 0 Multiply by 125: 25+10y+625=025 + 10y' + 625 = 0 10y=65010y' = -650 y=65y' = -65

So, the value of y(0)y'(0) is: y(0)=65y'(0) = -65

Would you like more details on any specific steps, or have any questions?


Follow-up Questions

  1. How would the process differ if we needed to find y(0)y''(0)?
  2. Can we analyze the behavior of y(x)y(x) around x=0x = 0 using this derivative?
  3. What if the equation were modified to xy10=0xy - 10 = 0; how would that affect the solution?
  4. How does implicit differentiation help in finding derivatives for complex equations?
  5. Can you explore the significance of y(0)y'(0) in terms of the graph of y=f(x)y = f(x)?

Tip

Using implicit differentiation is particularly useful for equations where solving explicitly for yy in terms of xx is challenging or impossible.

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Product Rule
Chain Rule
Calculus
Algebra

Formulas

Product rule: (uv)' = u'v + uv'
Chain rule: d/dx [f(g(x))] = f'(g(x)) * g'(x)

Theorems

Implicit Differentiation
Rules of Derivatives

Suitable Grade Level

Undergraduate Calculus (first-year college)