What is the Implicit Differentiation Formula for sin(xy) = x+y?

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Discussion Overview

The discussion revolves around finding the implicit differentiation formula for the equation $\sin(xy) = x + y$. Participants explore the steps involved in applying implicit differentiation to this equation, including the derivatives of both sides and the manipulation of terms to isolate $\frac{dy}{dx}$.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • Some participants assert that the derivative of $\sin(xy)$ is $y \cos(xy)$, but they seek clarification on how to apply this in the context of the equation $\sin(xy) = x + y$.
  • One participant proposes that the differentiation leads to the equation $\cos(xy)(y + xy') = 1 + y'$, which is a key step in the process.
  • Another participant elaborates on the steps to isolate $y'$ by distributing and rearranging terms, ultimately leading to the expression $\frac{y \cos(xy) - 1}{1 - x \cos(xy)} = y' = \frac{dy}{dx}$.
  • There is a repetition of the same steps by different participants, indicating a shared understanding of the process but also a potential for redundancy in the discussion.

Areas of Agreement / Disagreement

Participants generally agree on the steps involved in deriving the implicit differentiation formula, but there is no explicit consensus on the final presentation of the formula, as some participants reiterate the same points without introducing new perspectives.

Contextual Notes

Some steps in the differentiation process may depend on the assumptions made about the functions involved, and the discussion does not resolve any potential ambiguities in the application of implicit differentiation.

Who May Find This Useful

This discussion may be useful for students or individuals seeking to understand implicit differentiation, particularly in the context of trigonometric functions involving multiple variables.

karush
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if $\displaystyle \sin{\left(xy\right)}=x+y$, then $\displaystyle\frac{dy}{dx}=$

know this is implicit differentiation and that $\displaystyle\frac{dy}{dx}$ of $\displaystyle\sin(xy)$ is $\displaystyle y\cos{(xy)}$ but how is this done with $= x + y$

the answer to this is
$
\displaystyle
\frac{y \cos{⁡(xy)}-1}{1-x \cos{(⁡xy)}}
$
 
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karush said:
if $\displaystyle \sin{\left(xy\right)}=x+y$, then $\displaystyle\frac{dy}{dx}=$

know this is implicit differentiation and that $\displaystyle\frac{dy}{dx}$ of $\displaystyle\sin(xy)$ is $\displaystyle y\cos{(xy)}$ but how is this done with $= x + y$

the answer to this is
$
\displaystyle
\frac{y \cos{⁡(xy)}-1}{1-x \cos{(⁡xy)}}
$
[math]\frac{ d(sin(xy)) }{dx} = cos(xy) \times \frac{d(xy)}{dx}[/math]

[math]\frac{d( sin(xy) ))}{dx} = cos(xy) \times (y + xy')[/math]

Can you do it from here?

-Dan
 
topsquark said:
[math]\frac{d( sin(xy) ))}{dx} = cos(xy) \times (y + xy')[/math]

Can you do it from here?

-Dan

$\displaystyle\frac{dy}{dx}(x+y)= 1+y'$

so,

$\cos{xy}\left[y+xy'\right]
=1+y'
$

so, distributing and isolating terms with y'

$\displaystyle
y\cos(xy)-1=y' -x\cos(xy)y' = (1-x\cos(xy))y'
$

dividing

$\displaystyle
\frac{y\cos(xy)-1}{1-x\cos(xy)} = y'=\frac{dy}{dx}
$
 
Last edited:
karush said:
$\displaystyle\frac{dy}{dx}(x+y)= 1+y'$

so,

$\cos{xy}\left[y+xy'\right]
=1+y'
$

so, distributing and isolating terms with y'

$\displaystyle
y\cos(xy)-1=y' -x\cos(xy)y' = (1-x\cos(xy))y'
$

dividing

$\displaystyle
\frac{y\cos(xy)-1}{1-x\cos(xy)} = y'=\frac{dy}{dx}
$
Looks good to me. :)

-Dan
 

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