Thread: [SOLVED] Implicit Differentiation

$\displaystyle y^2=cos(xy)+ln(x+y)$

Find $\displaystyle \frac{dy}{dx}$ in terms of $\displaystyle x$ and $\displaystyle y$


Now, this is what I did, I'm not entirely sure if I'm doing it right.

$\displaystyle 2y.\frac{dy}{dx} = -ysin(xy).(y+x\frac{dy}{dx})+\frac{1}{x} + \frac{1}{y}\frac{dy}{dx}$

$\displaystyle \frac{-1}{x}+ysin(xy)=(\frac{1}{y}-xsin(xy)-2y)\frac{dy}{dx}$

$\displaystyle \frac{dy}{dx}=\frac{ysin(xy)-\frac{1}{x}}{\frac{1}{xy}-xsin(xy)-2y}$

So do i get or ?

2y*dy=-sin(xy)*(y+x*dy)+(1+dy)/(x+y)
where dy=dy/dx
solve dy you'll get the answer?

Originally Posted by Rapid_W

$\displaystyle y^2=cos(xy)+ln(x+y)$

Find $\displaystyle \frac{dy}{dx}$ in terms of $\displaystyle x$ and $\displaystyle y$


Now, this is what I did, I'm not entirely sure if I'm doing it right.

$\displaystyle 2y.\frac{dy}{dx} = {\color{blue}-ysin(xy).(y+x\frac{dy}{dx})}+{\color{red}\frac{1}{ x} + \frac{1}{y}\frac{dy}{dx}}$

$\displaystyle \frac{-1}{x}+ysin(xy)=(\frac{1}{y}-xsin(xy)-2y)\frac{dy}{dx}$

$\displaystyle \frac{dy}{dx}=\frac{ysin(xy)-\frac{1}{x}}{\frac{1}{xy}-xsin(xy)-2y}$

So do i get or ?

can you find your mistake in the blue and the red term or
want I tell you about them

Not really sure about the blue, but is the red supposed to be

$\displaystyle \frac{1}{xy}\frac{dy}{dx}$?

$\displaystyle ln(x+y)...derive...\frac{1}{x+y}+\frac{dy/dx}{y+x} $

ahh ok, should the blue be this?

$\displaystyle -ysin(xy)-xsin(xy).\frac{dy}{dx}$

yeah that's right you are correct

ok so starting over doing it correctly i get

$\displaystyle 2y\frac{dy}{dx} = -sin(xy).y - sin(xy).x\frac{dy}{dx} + \frac{1}{x+y} + \frac{1}{x+y}\frac{dy}{dx}$
$\displaystyle
(2y + xsin(xy) - \frac{1}{x+y})\frac{dy}{dx} = -ysin(xy) + \frac{1}{x+y}$

giving me an answer of

$\displaystyle \frac{dy}{dx} = \frac{-ysin(xy)+(x+y)^{-1}}{2y+xsin(xy)-(x+y^{-1}}$

which simplifys to

$\displaystyle \frac{dy}{dx} = \frac{-ysin(xy)}{2y+xsin(xy)} -1$

Is this correct?

Originally Posted by Rapid_W

ok so starting over doing it correctly i get

$\displaystyle 2y\frac{dy}{dx} = -sin(xy).y - sin(xy).x\frac{dy}{dx} + \frac{1}{x+y} + \frac{1}{x+y}\frac{dy}{dx}$
$\displaystyle
(2y + xsin(xy) - \frac{1}{x+y})\frac{dy}{dx} = -ysin(xy) + \frac{1}{x+y}$

giving me an answer of

$\displaystyle \frac{dy}{dx} = \frac{-ysin(xy)+(x+y)^{-1}}{2y+xsin(xy)-(x+y^{-1}}$

which simplifys to

$\displaystyle \frac{dy}{dx} = \frac{-ysin(xy)}{2y+xsin(xy)} -1$

Is this correct?

$\displaystyle \frac{dy}{dx} = \frac{-ysin(xy)+(x+y)^{-1}}{2y+xsin(xy)-(x+y)^{-1}}$ this is correct

but how that became to this

$\displaystyle \frac{dy}{dx} = \frac{-ysin(xy)}{2y+xsin(xy)}-1 $

I am not sure if it is wrong or not

$\displaystyle \frac{dy}{dx} = \frac{-ysin(xy)+(x+y)^{-1}}{2y+xsin(xy)-(x+y)^{-1}}$



$\displaystyle \frac{dy}{dx} =\left(\dfrac{ \dfrac{(x+y)(-ysin(xy))+1 }{(x+y)} }{\dfrac{ (x+y)(2y+xsin(xy)-1)}{(x+y)}}\right)$




$\displaystyle \frac{dy}{dx} =\left(\dfrac{ (x+y)(-ysin(xy))+1 }{(x+y)(2y+xsin(xy)-1)}\right)$

simplify it

ahh ok, that makes more sense, ok, my problem is now solved thanks everyone that helped