Use implicit differentiation to find dy/dx

In summary, the student attempted to solve the equation using implicit differentiation, but got it wrong. They were able to solve the equation using the chain rule after referencing the previous problem.
  • #1
domyy
196
0

Homework Statement



Use implicit differentiation to find dy/dx.

Homework Equations



xey - 10x + 3y = 0

The Attempt at a Solution



= [xey + ey(y)'] - (10x)' + (3y)' = 0

= xey + ey(y') - 10 + 3(y') = 0

= y' (ey + 3) = 10 - xey

= y' = 10 - xey/ ey + 3y

However, my book says the answer: 10 - ey/ xey + 3y.

What am I doing wrong?

Thanks!
 
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  • #2
domyy said:

Homework Statement



Use implicit differentiation to find dy/dx.

Homework Equations



xey - 10x + 3y = 0

The Attempt at a Solution

The first thing wrong is starting the line below with =. By putting that = at the start of each line, you are more likely to lose track of the two sides of the equation you started with.
domyy said:
= [xey + ey(y)'] - (10x)' + (3y)' = 0
The more serious mistake is above. You didn't take the derivative of xey correctly. d/dx(xey) = ey + xey*y'. Do you see why?
domyy said:
= xey + ey(y') - 10 + 3(y') = 0

= y' (ey + 3) = 10 - xey

= y' = 10 - xey/ ey + 3y

However, my book says the answer: 10 - ey/ xey + 3y.
No, I believe their answer is (10 - ey)/(xey + 3). Note the parentheses, and the 3 instead of the 3y that you had.
domyy said:
What am I doing wrong?

Thanks!
 
  • #3
About taking the derivative wrongly, my question is: How does that differ from y= xe^x - e^x?

Because I was trying to follow the same thinking when solving the problem in question.

This is how I solved it ( and I got it right):

= xe^x + e^x (x)' - e^x(x)'
= xe^x + e^x - e^x
= e^x ( x -1 + 1)
= xe^x

Thanks!
 
Last edited:
  • #4
domyy said:
About taking the derivative wrongly, my question is: How does that differ from y= xe^x - e^x?

Because I was trying to follow the same thinking when solving the problem in question.

This is how I solved it ( and I got it right):
Don't start your work with =.
domyy said:
= xe^x + e^x (x)' - e^x(x)'
= xe^x + e^x - e^x
= e^x ( x -1 + 1)
= xe^x

Thanks!
In this problem,
d/dx(xex) = x *d/dx(ex) + d/dx(x) * ex = xex + 1*ex = ex(x + 1)

The previous problem is different because you need to use the chain rule.
d/dx(xey) = x * d/dx(ey) + d/dx(x) * ey
= x * d/dy(ey) * dy/dx + d/dx(x) * ey = xey*y' + 1*ey = ey(xy' + 1)
 
  • #5
Ok, I am going to try to solve some other problems to practice it.
 
Last edited:

1. What is implicit differentiation?

Implicit differentiation is a method used in calculus to find the derivative of a function that is defined implicitly, meaning it is not explicitly written in terms of one variable. This is often the case for equations involving multiple variables or complex functions.

2. How is implicit differentiation different from explicit differentiation?

Explicit differentiation involves finding the derivative of a function that is written explicitly in terms of one variable. Implicit differentiation, on the other hand, is used for functions that are not explicitly written in terms of one variable and requires the use of the chain rule.

3. What is the process for using implicit differentiation to find dy/dx?

The process for implicit differentiation involves taking the derivative with respect to x on both sides of the equation, using the chain rule when necessary, and isolating dy/dx on one side of the equation. Then, solve for dy/dx to get the derivative of the function.

4. When should implicit differentiation be used?

Implicit differentiation should be used when the function is not explicitly written in terms of one variable, making it difficult or impossible to find the derivative using explicit differentiation. This method is commonly used in physics, engineering, and other fields that involve complex equations.

5. What are some common mistakes to avoid when using implicit differentiation?

Some common mistakes to avoid when using implicit differentiation include forgetting to apply the chain rule, not isolating dy/dx, and misplacing negative signs. It is important to carefully follow the steps of implicit differentiation and check your work to avoid these errors.

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