Question #0ed82

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Answer 1 · 2016-11-19T22:31:23

$y'(5)=-16/5$

$d/dx(3x^2+3x+xy) = d/dx5$

$=> (d/dx3x^2)+(d/dx3x)+(d/dxxy) = 0$

$=> 6x + 3 + [x(d/dxy) + y(d/dxx)] = 0$

$=> 6x + 3 + xdy/dx + y = 0$

$=> dy/dx = -(6x+y+3)/x$

Evaluated at $x=5$ , and noting that $x=5$ implies $y=y(5) = -17$ , we have

$y'(5) = dy/dx|_(x=5)$

$= -(6(5)+(-17)+3)/(5)$

$=-(30-17+3)/5$

$=-16/5$