Implicit Differentiation of 5=2x^2y+7xy^2

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Discussion Overview

The discussion revolves around finding the derivative \( \frac{dy}{dx} \) of the equation \( 5 = 2x^2y + 7xy^2 \). Participants explore the process of implicit differentiation, addressing the steps involved and clarifying misunderstandings related to the differentiation of constants and terms involving both \( x \) and \( y \).

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants express confusion about how to differentiate the constant \( 5 \), with one noting that the derivative of a constant is \( 0 \).
  • There are multiple attempts to apply implicit differentiation, with varying expressions for \( \frac{dy}{dx} \) being proposed.
  • One participant suggests that the derivative of \( 2x^2y \) should include a term involving \( y' \), while another participant corrects the earlier expression, indicating that it should be \( 4xy + 2x^2y' \).
  • Participants engage in correcting each other's work, with some pointing out errors in signs and terms, particularly regarding the placement of negative signs in the derivatives.
  • Several participants arrive at similar forms for \( y' \), but the discussion remains focused on the steps taken to isolate \( y' \) and the correctness of the differentiation process.

Areas of Agreement / Disagreement

There is no consensus on the correct approach to the differentiation, as participants present competing views and corrections. The discussion remains unresolved with respect to the final expression for \( \frac{dy}{dx} \>.

Contextual Notes

Participants highlight potential errors in differentiation steps, particularly in the handling of terms involving \( y' \) and the differentiation of products. The discussion reflects a range of interpretations and corrections without settling on a definitive method or result.

karush
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find dy/dx of
$$5=2x^2y+7xy^2$$

ok didn't know how to set this up with the $5=$
 
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karush said:
find dy/dx of
$$5=2x^2y+7xy^2$$

ok didn't know how to set this up with the $5=$
When you take the derivative of both sides you wind up with [math]\dfrac{d}{dx}(5)[/math]. What is the derivative of a constant?

-Dan
 
5x

But how do we know it is x not y
 
karush said:
5x

But how do we know it is x not y
You are trying to find the derivative [math]\dfrac{dy}{dx}[/math] right?

[math]5 = 2x^2 y + 7x y^2[/math]

[math]\dfrac{d}{dx}(5) = \dfrac{d}{dx} (2x^2 y) + \dfrac{d}{dx} (7x y^2) [/math]

-Dan
 
karush said:
5x

But how do we know it is x not y
NO! You appear to be integrating rather than differentiating. The derivative of any constant is 0.
 
$\displaystyle \dfrac{d}{dx}(5) = \dfrac{d}{dx} (2x^2 y) + \dfrac{d}{dx} (7x y^2)$
d/dx
$0=4xy-2x^2yy'+7y^2-14xyy'$
isolate y'
$-2x^2yy'-14xyy'=4xy+7y^2$
$y'(-2x^2y-14xy)=4xy+7y^2$
simplify
$\displaystyle y'= \frac{(4 xy + 7 y^2)}{(-2x^2 - 14x y)}=- \frac{y (4 x + 7 y)}{2 x (x + 7 y)}$
 
Last edited:
karush said:
$\displaystyle \dfrac{d}{dx}(5) = \dfrac{d}{dx} (2x^2 y) + \dfrac{d}{dx} (7x y^2)$
d/dx
$0=4xy {\color{red}{+} 2x^2 y'}+7y^2 {\color{red}{+}} 14xyy'$

corrections ... try again
 
done try more!
$0=4xy-2x^2y'+7y^2-14xyy'$
isolate y'
$-2x^2y'-14xyy'=4xy+7y^2$
$y'(-2x^2-14xy)=4xy+7y^2$
simplify
$\displaystyle
y'= \frac{(4 xy + 7 y^2)}{(-2x^2 - 14x y)}
=- \frac{y (4 x + 7 y)}{2 x (x + 7 y)}$
 
karush said:
done try more!
$0=4xy-2x^2y'+7y^2-14xyy'$
isolate y'
$-2x^2y'-14xyy'=4xy+7y^2$
$y'(-2x^2-14xy)=4xy+7y^2$
simplify
$\displaystyle
y'= \frac{(4 xy + 7 y^2)}{(-2x^2 - 14x y)}
=- \frac{y (4 x + 7 y)}{2 x (x + 7 y)}$
Look more carefully at skeeter's post.
[math]\dfrac{d}{dx}(2x^2 y) = 4xy + 2x^2y'[/math]

[math]\dfrac{d}{dx}(7x y^2) = 7y^2 + 14xy y'[/math]

Why are you putting the negative sign in there?

-Dan
 
  • #10
like this
$0=4xy+2x^2y'+7y^2+14xyy'$
isolate y'
$-2x^2y'-14xyy'=4xy+7y^2$
$y'(-2x^2-14xy)=4xy+7y^2$
simplify
$\displaystyle
y'= \frac{(4 xy + 7 y^2)}{(-2x^2 - 14x y)}
=- \frac{y (4 x + 7 y)}{2 x (x + 7 y)}$
 

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