How do you differentiate this equation implicitly with respect to x: xy + 21 = 0?

I made this example up. There's a problem in my book just like it.

2012-11-12T07:50:37Z

Please explain why you do what you do, thanks.

BrainyMe2012-11-12T08:00:45Z

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Since you're differentiating implicitly with respect to x, this calls for you to use he chain rule on the term 'xy'. Obviously the derivatives of 21 and 0 will be 0.
First take the derivative with respect to x and that leaves you with y.
Secondly, since we are differentiating with respect to x still take the derivative of y as usual but you will multiply the derivative by this term- dy/dx. So the derivative of the y in the equation would be x*(dy/dx)
You add up both derivatives as you know how to do with the chain rule, and you have the expression below:
y +x(dy/dx) = 0
You solve for dy/dx from there

Anonymous2016-11-25T01:01:51Z

differentiate the two sides: d/dx [xy - x - 5y - 19] = d/dx [0] d/dx [xy] - d/dx [x] - 5 d/dx [y] - d/dx [19] = 0 x d/dx [y] + y d/dx [x] - a million - 5(dy/dx) - 0 = 0 -- product rule x(dy/dx) + y(a million) - 5(dy/dx) = a million (dy/dx)(x - 5) = a million-y dy/dx = (a million-y) / (x-5)

Ed I2012-11-12T07:56:11Z

xy = -21
x dy/dx + y = 0
x dy/dx = -y
dy/dx = -y/x