How do you differentiate $e^{y} cos (x) = 6 + sin (x y)$ $e^y cos(x) = 6 + sin(xy)$ ?

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Answer 1 · 2018-04-17T15:39:08

$\frac{d y}{d x} = \frac{e^{y} sin x + y cos (x y)}{e^{y} cos x - x cos (x y)}$ $(dy)/(dx)=(e^ysinx+ycos(xy))/(e^ycosx-xcos(xy)]$

Here,

$e^{y} cos x = 6 + sin (x y)$ $e^ycosx=6+sin(xy)$

Diff.w.r.t. $x$ $x$ , $using product and chain rule$ $"using "color(blue)"product and chain rule"$

$e^{y} \frac{d}{d x} (cos x) + cos x \frac{d}{d x} (e^{y}) = 0 + cos (x y) \frac{d}{d x} (x y)$ $e^yd/(dx)(cosx)+cosxd/(dx)(e^y)=0+cos(xy)d/(dx)(xy)$

$e^{y} (- sin x) + cos x (e^{y} \frac{d y}{d x}) = cos (x y) [x \frac{d y}{d x} + y \cdot 1]$ $e^y(-sinx)+cosx(e^y(dy)/(dx))=cos(xy)[x(dy)/(dx)+y*1]$

$- e^{y} sin x + e^{y} cos x \frac{d y}{d x} = x cos (x y) \frac{d y}{d x} + y cos (x y)$ $-e^ysinx+e^ycosx(dy)/(dx)=xcos(xy)(dy)/(dx)+ycos(xy)$

$e^{y} cos x \frac{d y}{d x} - x cos (x y) \frac{d y}{d x} = e^{y} sin x + y cos (x y)$ $e^ycosx(dy)/(dx)-xcos(xy)(dy)/(dx)=e^ysinx+ycos(xy)$

$\frac{d y}{d x} [e^{y} cos x - x cos (x y)] = e^{y} sin x + y cos (x y)$ $(dy)/(dx)[e^ycosx-xcos(xy)]=e^ysinx+ycos(xy)$

$\frac{d y}{d x} = \frac{e^{y} sin x + y cos (x y)}{e^{y} cos x - x cos (x y)}$ $(dy)/(dx)=(e^ysinx+ycos(xy))/(e^ycosx-xcos(xy)]$

How do you differentiate e^y cos(x) = 6 + sin(xy)eycos(x)=6+sin(xy)e^y cos(x) = 6 + sin(xy)?