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Use implicit differentiation to find dy/dx

  • Thread starter domyy
  • Start date
  • #1
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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!
 

Answers and Replies

  • #2
33,176
4,859

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.
= [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?
= 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.
What am I doing wrong?

Thanks!
 
  • #3
196
0
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
33,176
4,859
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 =.
= 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
196
0
Ok, I am going to try to solve some other problems to practice it.
 
Last edited:

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