Implicit differentiation - messed up somewhere

In summary: At (1, 1), we have 1+ 2+ 2= 5 so d^2y/dx^2= 1/5.In summary, using implicit differentiation, we can find the second derivative, d^2y/dx^2, at (1,1) for the equation x^2y + x^2 -y^2 = 1. After finding the first derivative, dy/dx, we can use implicit differentiation again to find d^2y/dx^2, which is equal to 1/5 at (1,1).
  • #1
meredith
16
0

Homework Statement


find d^2y/dx^2 at (1,1) for:
x^2y + x^2 -y^2 = 1




Homework Equations


none



The Attempt at a Solution


i worked it all out but the answer I am getting is not an option. could someone show me where i made a mistake? I am not asking you to do the problem for me, just fix my error. this is what i did:
first i found dy/dx:

x^2(dy/dx) + y(2x) +2x -2y(dy/dx)=0
(x^2 - 2y)(dy/dx) = -2xy - 2x
dy/dx = (-2xy-2x)/(x^2-2y)
when i put (1,1) in, i got 4

then to find the second derivative i used the quotient rule:
d^2y/dy^2 = [(x^2-2y) (-2x)(dy/dx) +y(-2) -2] - ((-2xy-2x)(2x-2)(dy/dx)]/(x^2-2y)^2

then i just put the values in. for the x's and y's i put in 1, and for dy/dx i put in 4. i ended up with:
[(-1 x -2 x 4 + -2 -2) - (-2 -2) x (2-2(4) ] / 1 = 28 ? but this answer is incorrect. what did i do wrong?


Thanks so much!
 
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  • #2
It's not necessary to solve for dy/dx- just use implicit differentiation again.

You are correct that x^2(dy/dx) + y(2x) +2x -2y(dy/dx)= 0.

Now, (x^2(dy/dx)'= x^2 d^2y/dx^2+ 2x dy/dx, (2xy)'= 2xy'+ 2y, and (2y dy/dx)'= 2y d^2y/dx^2+ 2 (dy/dx)^2. Put those together.
 

1. What is implicit differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of a function that is expressed implicitly, rather than explicitly. In other words, it is used to find the rate of change of a dependent variable with respect to an independent variable when the equation of the function is not explicitly given in terms of the variables.

2. How is implicit differentiation different from explicit differentiation?

Explicit differentiation involves finding the derivative of a function where the equation is given explicitly in terms of the variables. Implicit differentiation, on the other hand, is used when the equation is not given explicitly and involves treating one of the variables as an implicit function of the other.

3. What are the steps for performing implicit differentiation?

The steps for implicit differentiation are as follows:

  1. Identify the dependent and independent variables in the equation.
  2. Take the derivative of both sides of the equation with respect to the independent variable.
  3. Use the chain rule to find the derivative of any terms that involve the dependent variable.
  4. Solve the resulting equation for the derivative of the dependent variable.

4. What are some common mistakes made when using implicit differentiation?

Some common mistakes when using implicit differentiation include:

  • Forgetting to use the chain rule when taking the derivative of terms involving the dependent variable.
  • Misidentifying the dependent and independent variables in the equation.
  • Not simplifying the resulting equation before solving for the derivative.
  • Not checking the final answer with the original equation to ensure accuracy.

5. How can I check my answer when using implicit differentiation?

You can check your answer by substituting the derivative back into the original equation and verifying that it satisfies the equation. You can also take the second derivative of the original function and see if it matches your calculated derivative.

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