If f = 0, then df/dt=0, but ∂f/∂x≠0

  • Thread starter xorg
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  • #1
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Suppose a function f (t) = 0
x (t) -y (t) = 0
with
x = t
y = t
df/dt = 0
However
∂f/∂x = 1

This case may seem obvious to most of the regulars this forum, but took me by surprise when I was reading a math book that I needed to "derive from both sides" (I assumed that as a side was zero, or other would always be).

I wonder if there is any book that deals with these things in a more elegant way. To respond in a clear way, when I can "derive from both sides."

Because for example:
x (t) - y (t) = 0
"deriving from both sides":
∂x/∂x - ∂y/∂x = 0
1 = 0

Many thanks for the help, and sorry if my question is very simple.
 

Answers and Replies

  • #2
FactChecker
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I assumed that as a side was zero, or other would always be.
Both sides ARE zero, but the left side maintains zero by trading off the two terms.
I wonder if there is any book that deals with these things in a more elegant way. To respond in a clear way, when I can "derive from both sides."
This doesn't require an entire book (although there may be one). This is called "implicit differentiation". The equation defines an implicit relationship between x and y. In your example, you could solve for one explicitly in terms of the other, but other examples are much more difficult to solve explicitly. So implicit differentiation is another way to solve the problem without explicitly solving the equation.
∂x/∂x - ∂y/∂x = 0
1 = 0
.
There is your problem. ∂y/∂x is not 0. This is where you find out that 1 - ∂y/∂x = 0, so ∂y/∂x = 1.
 
  • #3
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I was reading a math book that I needed to "derive from both sides"
In English, we call it "differentiating" not "deriving".
 

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