3.7.3-eg-1 - Example Use implicit differentiation to find...

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Unformatted text preview: Example Use implicit differentiation to find y (x) if (x, y (x)) satisfy x2 + 3xy + y 2 = 11, and then find the equation of the tangent line to this curve at the point (1 , 2). Solution: As a reminder that y depends on x, the above equation can be written x2 + 3xy (x) + (y (x))2 = 11. Differentiating both sides of the equation with respect to x, d d (11), x2 + 3xy (x) + (y (x))2 = dx dx leads to 2x + 3x · y (x) + 3y (x) + 2y (x) · y (x) = 0, (1) where we have used the product rule to evaluate d (3x · y (x)) = (3x) · (y (x)) + (3x) · y (x) = 3xy (x) + 3y (x), dx and the Chain Rule to derive d (y (x))2 = 2y (x) · y (x). dx In order to solve equation (1) for y (x), collect all terms involving y (x) on the left-hand side of the equation, and all other terms on the right-hand side. That is, 3x · y (x) + 2y (x) · y (x) = −2x − 3y (x), or, combining terms and replacing y (x) and y (x) by their simpler expressions y and y , respectively, (3x + 2y )y = −2x − 3y, (2) an equation of the form A(x, y )y = B (x, y ), where A(x, y ) = 3x + 2y, and B (x, y ) = −2x − 3y. Solving equation (2) for y results in y= −2x − 3y B (x, y ) = . A(x, y ) 3x + 2y (3) The slope of the tangent line to the curve at the point (1, 2) is given by y in (3) where the x and y values in (3) are given by 1 and 2, respectively. It follows that y (1) = −2 · 1 − 3 · 2 8 =− . 3·1+2·2 7 So the equation of the tangent line to the curve x2 + 3xy + y 2 = 11 at the point (1, 2) is given by 8 y − 2 = − (x − 1). 7 2 ...
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3.7.3-eg-1 - Example Use implicit differentiation to find...

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