Math Problem Statement

Using implicit differentiation, (Second Derivative), Knowing that now, solve If xy-y= 4x then find the 2nd derivative at the point (3,6) in simplest form. Derivate with respect to x, use chain rule for y terms, rearrange terms with y prime on the left, other terms on the right, then divide, simplify any common factors, convert y prime to dy/dx, then find the 2nd derivative using the product or quotient rule, then plug in y, y prime, and x. You neep to apply the point to the 2nd derivative NOT the first

Solution

To find the second derivative of the equation xyy=4xxy - y = 4x at the point (3,6)(3, 6) using implicit differentiation, we follow these steps:

Step 1: Implicit Differentiation

Start with the equation: xyy=4xxy - y = 4x Differentiating both sides with respect to xx: ddx(xy)ddx(y)=ddx(4x)\frac{d}{dx}(xy) - \frac{d}{dx}(y) = \frac{d}{dx}(4x)

Using the product rule on xyxy: y+xdydxdydx=4y + x\frac{dy}{dx} - \frac{dy}{dx} = 4 Rearranging terms gives: y+(x1)dydx=4y + (x - 1)\frac{dy}{dx} = 4 Now isolate dydx\frac{dy}{dx}: (x1)dydx=4y(x - 1)\frac{dy}{dx} = 4 - y dydx=4yx1\frac{dy}{dx} = \frac{4 - y}{x - 1}

Step 2: Convert to dy/dxdy/dx

This can be rewritten as: y=4yx1y' = \frac{4 - y}{x - 1}

Step 3: Find the First Derivative at Point (3, 6)

Substituting x=3x = 3 and y=6y = 6: y=4631=22=1y' = \frac{4 - 6}{3 - 1} = \frac{-2}{2} = -1

Step 4: Second Derivative

Now we differentiate y=4yx1y' = \frac{4 - y}{x - 1} again using the quotient rule: d2ydx2=(x1)(dydx)(4y)(1)(x1)2\frac{d^2y}{dx^2} = \frac{(x - 1)(-\frac{dy}{dx}) - (4 - y)(1)}{(x - 1)^2} Substituting y=1y' = -1 into the expression: d2ydx2=(x1)((1))(4y)(1)(x1)2\frac{d^2y}{dx^2} = \frac{(x - 1)(-(-1)) - (4 - y)(1)}{(x - 1)^2} At x=3x = 3 and y=6y = 6: d2ydx2=(31)(1)(46)(1)(31)2=2+24=44=1\frac{d^2y}{dx^2} = \frac{(3 - 1)(1) - (4 - 6)(1)}{(3 - 1)^2} = \frac{2 + 2}{4} = \frac{4}{4} = 1

Final Result

The second derivative d2ydx2\frac{d^2y}{dx^2} at the point (3,6)(3, 6) is: 1\boxed{1}

Summary and Questions

If you have any further questions or need more details, please let me know! Here are some related questions you might consider:

  1. How would you find the second derivative using a different implicit function?
  2. Can you explain how the product rule works in implicit differentiation?
  3. What are the applications of second derivatives in real-world problems?
  4. How do you interpret the significance of the second derivative at a point?
  5. What would happen if the point were different, say (2,3)(2, 3)?

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Second Derivatives
Differential Calculus

Formulas

dy/dx = (4 - y) / (x - 1)
d^2y/dx^2 = [(x - 1)(dy/dx) - (4 - y)] / (x - 1)^2

Theorems

Product Rule
Quotient Rule

Suitable Grade Level

Grades 11-12