Setting Derivative = 0 and solving

  • Thread starter Thread starter deedsy
  • Start date Start date
  • Tags Tags
    Derivative
Click For Summary
Setting the derivative equal to zero to solve for a variable is valid, even when the derivative is a fraction with the variable in both the numerator and denominator. It is permissible to multiply both sides by the denominator, provided it does not equal zero, effectively simplifying the equation. Alternatively, dividing by x/x can also be used, assuming x is not zero. Both methods lead to the same solution, demonstrating the flexibility in approach to solving such problems. Understanding these concepts is crucial for tackling more complex equations in calculus.
deedsy
Messages
81
Reaction score
0

Homework Statement


I'm currently working on a problem that requires me to set the derivative = 0 and solve for a variable (call it x). The derivative comes out to be a fraction, with x terms in both the numerator and denominator. Is it legal to just multiply 0 by the denominator (thereby canceling it) even if it has the term of interest as part of it?

Simple Ex: say the derivative came out to be x-3 / 2x. And I want to solve for x.
When I set that derivative equal to zero, can i just multiply 0 by 2x, leaving x-3=0? So x=3

Homework Equations



none

The Attempt at a Solution


...
 
Physics news on Phys.org
That should be ok. You could also divide the left side by x / x, assuming that x is not equal to zero, resulting in:

( 1 - (3/x) ) / 2 = 0
 
  • Like
Likes 1 person
thank you
 
rcgldr said:
That should be ok. You could also divide the left side by x / x, assuming that x is not equal to zero, resulting in:

( 1 - (3/x) ) / 2 = 0

i thought they're supposed to show the work?
 
MGCLO said:
i thought they're supposed to show the work?

It is a requirement for HW problems, but my question was geared towards a concept. The equation I'm deriving for the HW would probably take up an entire line on here. The example I put on was just that, an example, it wasn't even close to my actual problem (although I wish it was haha)
 
MGCLO said:
i thought they're supposed to show the work?
Multiplying both sides by 2x is just as valid as dividing the left side by x/x. I only showed that as an alternative in case there's a situation where that would be a better option for a different equation.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K