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anonymous
 5 years ago
could anyone explain to me why we attach (dy/dx) to y when using implicit differentiation? I know it has something to do with the chain rule
anonymous
 5 years ago
could anyone explain to me why we attach (dy/dx) to y when using implicit differentiation? I know it has something to do with the chain rule

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let's consider a simple case, and then apply the chain rule: y = sin(xy^3) dy/dx = derivative of inside * derivative of outside dy/dx = d/dx(xy^3) * d/dx(sin(xy^3)) So, we have a product in the first term: x * derivative of y^3 + y^3 * derivative of x. dy/dx = (3 * x* y^2 * dy/dx + y^3)*cos(xy^3). The extra dy/dx is because the derivative of y is dy/dx, assuming that y is the function itself. So, say you want to differentiate y^2 with respect to x: it's actually just a case of the chain rule. (y)^2. Derivative of inside * derivative of outside. Hence, dy/dx * 2 * y.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok maybe a heuristic will help me, when using implicit differentiation (say x^2+y^2=1) you simply derive y^2 and attach the dy/dx always? Could that go wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think it might go wrong in a question here or there; it's always safer to remember that in problems like this, the derivative of y is dy/dx, and you'll be fine. For \[x^2+y^2=1\] you differentiate as such:\[2x + 2 * y * \frac{dy}{dx} = 0\] because you use the chain rule on y^2  treat it as (y)^2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahh ok I got it now...thanks
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