derivative of a trig function with 2 variables

**AlphaTauLambda** · May 8th 2011, 02:32 PM

find dy/dx of sin(xy)=x+y

So far I have

cos(xy)*(xy'+1y)=1+y'
(cos(xy))xy'+(cos(xy))y=1+y'

Then, I'm sure you solve here for y but for the life of me I don't see the logical steps to get to y.

Would it be

y=((1+y')-(cos(xy)xy'))/(cos(xy))

???

**skeeter** · May 8th 2011, 02:34 PM

Originally Posted by AlphaTauLambda

find dy/dx of sin(xy)=x+y

So far I have

cos(xy)*(xy'+1y)=1+y'
(cos(xy))xy'+(cos(xy))y=1+y'

Then, I'm sure you solve here for y but for the life of me I don't see the logical steps to get to y.

Would it be

y=((1+y')-(cos(xy)xy'))/(cos(xy))

???

solve for y', which is dy/dx ... not y

**AlphaTauLambda** · May 8th 2011, 02:51 PM

Skeeter. Thanks. I go to school in North Texas... haha

The person who helped me with this problem has written (xcosxy-y)y'=1-ycosxy

I don't understand the transition from the previous step (which is (cosxy)xy' + (cosxy)*y=1+y'

Any ideas?

**skeeter** · May 8th 2011, 02:59 PM

$\dfrac{d}{dx}\left[\sin(xy) = x+y]$

$\cos(xy) \cdot (xy' + y) = 1 + y'$

distribute the $\cos(xy)$ on the left ...

$xy'\cos(xy) + y\cos(xy) = 1 + y'$

get the terms w/ $y'$ on the same side ...

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

factor out $y'$ ...

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

$y' = \dfrac{1 - y\cos(xy)}{x\cos(xy) - 1}$

**AlphaTauLambda** · May 8th 2011, 03:10 PM

Thanks a million, Skeeter! This makes perfect sense!!!

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