Second Derivative using Implicit Differentiation

In summary, the problem is to find y''(x) of the parametric equation 9x^2+y^2=9 using implicit differentiation. The attempted solution includes finding y'(x) and using the quotient rule, but the result did not match the answer provided in the book. After using a substitution and simplifying the equation, the correct answer for y''(x) is -81/(y^3). The tip to look for substitutions and simplify the result was instrumental in solving this problem.
  • #1
Bear_B
18
0

Homework Statement


Find y''(x) of the parametric equation 9x^2+y^2=9 using implicit differentiation.

Homework Equations


I already came up with y'(x) = -9x/y

The Attempt at a Solution


Here is what I have for y''(x) so far

y''(x) = d/dx (-9xy^-1)
=-9(d/dx)(xy^-1)
=-9(x(d/dx)(y^-1)+(y^-1)(dx/dx))
=-9(-x(y^-2)y'(x)+y^-1)

I substituted the value of y'(x) = -9x/y here

=-9((-x/(y^2))(-9x/y)+(1/y))
=-9((9x^2)/(y^3)+(1/y))
=(-81x^2)/(y^3)-(9/y)

I know this is incorrect, I originally tried this using the quotient rule but was getting the same answer and the work was much more jumbled, so I opted for the product rule.

The book states the answer is -81/(y^3). I am stuck and haven't been able to work towards the right answer.
 
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  • #2
Use the quotient rule after you take the first derivative, then look for any possible substitutions that you think might work (& don't be afraid of fractions on fractions on fractions lol).

When you've got it as clean as you can get it (it isn't a messy equation but be careful w/ minus signs) look to see if multiplying the equation (i.e. top & bottom) by a clever choice of 1 will help.

let us know how it goes.
 
  • #3
You're good so far. Put your result over a common denominator and simplify the top.
 
  • #4
Ok, I got the answer. The tip on substitution was what really allowed my to break this one open. Since I think this is a great problem that requires a creative strategy (and possibly multiple attempts) to get the simplest answer, I am going to post the rest of my work here.

The result of the Product or Quotient Rule and the substitution for y'(x) yields:

y''(x)=(-9y-((81x^2)/y))/(y^2)

From the original equation, x^2=1-(y^2)/9
I used this value to substitute for x^2.

=(-9y-81((1-(y^2)/9)/(y^2)))/(y^2)

There are a lot of fractions going on, but from here it is just a matter of reducing and simplification.

=(-9y-81(1/y-(y^2)/9y))/(y^2)
=(-9y-81/y+81y/9)/(y^2)
=(-9y-81/y+9y)/(y^2)
=(-81/y)/(y^2)
=-81/y*1/(y^2)
y''(x)=-81/(y^3)

Thanks for the prompt tips. Your feedback helped me solve this problem and find the simplest way to express y''(x).
=
 

Related to Second Derivative using Implicit Differentiation

1. What is the second derivative using implicit differentiation?

The second derivative using implicit differentiation is a mathematical concept used to find the rate of change of the first derivative of a function where the independent variable is not explicitly stated in terms of the dependent variable. It is a way to find the curvature of a graph at a specific point.

2. How is the second derivative using implicit differentiation calculated?

To calculate the second derivative using implicit differentiation, first find the first derivative of the function using the power rule or chain rule. Then, differentiate the first derivative with respect to the independent variable, treating the dependent variable as a function of the independent variable. Finally, simplify the resulting equation to find the second derivative.

3. What is the purpose of finding the second derivative using implicit differentiation?

The second derivative using implicit differentiation allows us to determine the concavity of a graph at a specific point. This is useful in optimization problems, where we need to find the maximum or minimum value of a function. It is also used in curve sketching to identify points of inflection.

4. When is implicit differentiation used instead of explicit differentiation?

Implicit differentiation is used when the dependent variable cannot be explicitly written in terms of the independent variable, or when it is easier to differentiate the function implicitly rather than explicitly. This often occurs with curves and equations that cannot be easily solved for y in terms of x.

5. What are some real-life applications of the second derivative using implicit differentiation?

The second derivative using implicit differentiation is used in fields such as physics, engineering, and economics to analyze rates of change and optimize systems. It has applications in motion and acceleration, optimization of production processes, and identifying changes in supply and demand. It is also used in computer science to improve algorithms and optimize code.

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