Implicit Differentiation Question

Click For Summary
SUMMARY

The discussion centers on finding dy/dx through implicit differentiation for the equation e^(x^2 * y) = x + y. The user initially attempted to solve the equation using logarithmic properties but encountered issues with an online software marking their solution incorrect. The correct approach involved substituting x + y with e^(x^2 * y), which resolved the problem. This highlights the importance of proper notation and substitution in implicit differentiation.

PREREQUISITES
  • Understanding of implicit differentiation
  • Familiarity with exponential functions and their properties
  • Knowledge of logarithmic functions and their applications
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the principles of implicit differentiation in calculus
  • Learn about the properties of exponential and logarithmic functions
  • Explore common pitfalls in solving implicit equations
  • Practice additional problems involving implicit differentiation
USEFUL FOR

Students studying calculus, particularly those focusing on implicit differentiation, as well as educators seeking to clarify common mistakes in solving such equations.

RoyalFlush100
Messages
56
Reaction score
2

Homework Statement


I am told to find dy/dx by implicit differentiation where:
e^(x^2 * y) = x + y

Homework Equations


The above equation and the ln of it.

The Attempt at a Solution


e^(x^2 * y) = x + y
(x^2 * y)ln(e) = ln(x+y)
x^2 * y = ln(x+y)
x^2(dy/dx) + y(2x) = 1/(x+y) * (1 + dy/dx)
(dy/dx)[x^2 - 1/(x+y)] = 1/(x+y) - 2xy
dy/dx = (1/(x+y) - 2xy)/(x^2 - 1/(x+y))

or here: https://postimg.org/image/3k5ygbkxt/

This was marked wrong (online software). It doesn't care about simplest form and it was entered properly. So, what did I do wrong?
 

Attachments

  • 20161106_141310.jpg
    20161106_141310.jpg
    30.1 KB · Views: 372
Physics news on Phys.org
I got the same result without using the logarithm, only with a single quotient, i.e. expanded by ##x+y##. Maybe the missing brackets in your linear notation led to the online error. Or it is expected to write ##e^{x^2y}## instead of ##x+y## in the solution.
 
  • Like
Likes   Reactions: FactChecker
fresh_42 said:
I got the same result without using the logarithm, only with a single quotient, i.e. expanded by ##x+y##. Maybe the missing brackets in your linear notation led to the online error. Or it is expected to write ##e^{x^2y}## instead of ##x+y## in the solution.

I just put it in replacing x+y with e^(x^2 * y) and it worked. Thanks!
 
  • Like
Likes   Reactions: FactChecker

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
3
Views
2K
Replies
4
Views
2K