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

[karush]

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)}}
$

 

[topsquark]

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)}}
$

\(\displaystyle \frac{ d(sin(xy)) }{dx} = cos(xy) \times \frac{d(xy)}{dx}\)

\(\displaystyle \frac{d( sin(xy) ))}{dx} = cos(xy) \times (y + xy')\)

Can you do it from here?

-Dan

 

[karush]

\(\displaystyle \frac{d( sin(xy) ))}{dx} = cos(xy) \times (y + xy')\)

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}
$

 

[topsquark]

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

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