What is the Second Derivative of Implicitly Differentiated Function?

In summary, to determine y'' when 5x^2 + 3y^2 = 4, the first derivative was found using the power rule and chain rule. Then, the second derivative was found using the quotient rule. After substituting the value of y' into the expression for y'', the equation was simplified by using the original equation to solve for x^2. The final answer is -20/9y^3.
  • #1
Lion214
10
0

Homework Statement



Determine y'' when 5x^2 + 3y^2 = 4.



The Attempt at a Solution



So I found the first derivative using the power rule and chain rule,

10x + 6yy' = 0

Which I then solved for y',

y' = -10x/6y = -5x/3y

Next I found the second derivative using quotient rule,

y'' = (-15y + 15xy')/(3y)^2

This is the part where I am lost, since all the multiple choice answers involve no y' nor is there a x.
I don't know how to get rid of the x and the y' in the equation. Any help will be appreciated as to simplifying the equation even further and as to how.
 
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  • #2
Substitute the form of y' you have into the expression for y'' and simplify.
 
  • #3
Thanks statdad!

so here's what I did,

I substituted y' with -5x/3y and got

y'' = (-15y + 15x(-5x/3y))/(3y)^2 = (-15y - 25x^2/y)/(3y)^2 = (-15y^2/y - 25x^2/y)/(9y^2) = (-15y^2-25x^2)/9y^3

But I still had to get rid of the x, so I used the original equation to solve for x^2, which was;

x^2 = (4-3y^2)/5

Which I then use to simplify even further,

y'' = (-15y^2 - 25((4 - 3y^2)/5))/9y^3 = ((-75y^2 - 100 + 75y)/5)/9y^3

In which the answer is -20/9y^3.

Thanks for the insight. For some reason I didn't see that.
 
  • #4
You are welcome. The little bit of ``extra'' work you've just gone through is handy to remember for these types of problems - I would state with near 100% certainty you'll see similar things in the future.
 

What is implicit differentiation?

Implicit differentiation is a technique used in calculus to find the derivative of a function that is not explicitly expressed in terms of one variable. It allows us to find the rate of change of a dependent variable with respect to an independent variable, even when the relationship between the two variables is not explicitly defined.

Why is implicit differentiation useful?

Implicit differentiation is useful because it gives us a way to find the derivative of a function without having to solve for one variable in terms of the other. This is particularly helpful when dealing with complex equations that cannot be easily manipulated or solved for a specific variable.

How is implicit differentiation different from explicit differentiation?

The main difference between implicit and explicit differentiation is that explicit differentiation involves finding the derivative of a function that is expressed in terms of one variable, while implicit differentiation deals with finding the derivative of a function that is expressed in terms of multiple variables.

What are the steps for performing implicit differentiation?

The steps for performing implicit differentiation are as follows:

  1. Differentiate both sides of the equation with respect to the independent variable.
  2. Apply the chain rule to any terms that contain functions of the independent variable.
  3. Solve for the derivative of the dependent variable.

What are some real-world applications of implicit differentiation?

Implicit differentiation has many real-world applications, including in physics, economics, and engineering. For example, it can be used to find the rate of change of a moving object's position or to optimize a company's production costs. It is also commonly used in economics to analyze supply and demand curves.

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