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anonymous
 5 years ago
using implicit differentiation find y 'for x23xy4y2=23 and find the equation of the tangent line of the graph at (1,2)
anonymous
 5 years ago
using implicit differentiation find y 'for x23xy4y2=23 and find the equation of the tangent line of the graph at (1,2)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The point of implicit differentiation is to get y' when it isn't possible to get y as a function of x. So, when we implicitly differentiate, we differentiate both sides with respect to x because we want y as a function of x. So: I'm assuming x2 = x² and y2 = y² \[x^{2}  3xy  4y^{2} = 23\] \[d/dx(x^{2}  3xy  4y^{2})=d/dx(23)\] \[2x  d/dx(3xy)  8y(dy/dx)=0\] Notice that the middle term is a product rule so and the product rule is the derivative of the first fuction * the second function + the first function * the derivative of the second function. \[d/dx(3xy) = 3y + 3x(dy/dx)\] So the equation is: \[2x  (3y + 3x(dy/dx))  8y(dy/dx) = 0\] Simplify: \[2x  3y  3x(dy/dx)  8y(dy/dx) = 0\] \[2x  3y = (3x + 8y)(dy/dx)\] \[dy/dx = m= (2x  3y) / (3x + 8y)\] The slope of the tangent line is the derivative so just plug in the point (1,2) in for x and y in the above equation and then use the same point to get it in y = mx + b form: \[y = (8/13)x + b\] Now solve this equation for b using the point (1,2) again and you'll get the equation for the tangent line.
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