How do you determine dy/dx if $y=x+tan(xy)$ ?

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Answer 1 · 2018-05-23T14:07:56

$dy/dx=frac{cos^2(xy)+y}{cos^2(xy)-x}$

we get by the chain rule:
$y'=1+1/cos^2(xy)(y+xy')$

$dy/dx = 1 + y/cos^2(xy) + dy/dx (x/cos^2(xy))$

$dy/dx - dy/dx(x/cos^2(xy)) = 1+ysec^2(xy)$

$dy/dx * (1-frac{x}{cos^2(xy)}) = 1+ysec^2(xy)$

$dy/dx = frac{1+ysec^2(xy)}{1 - xsec^2(xy)}$

Multiplying by $cos^2(xy)$ over $cos^2(xy)$ we get

$dy/dx=frac{cos^2(xy)+y}{cos^2(xy)-x}$

How do you determine dy/dx if y=x+tan(xy)y=x+tan(xy)?