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A curve is given implicitly by the equation. 2 3 y + 6 xy + 4 x 2 − 2 y = 5 . 5 x − 3 y + 19 = 0. Find an equation for the tangent to the curve at the point P ( − 2,1 ) . 4 y + 5 x + 6 = 0. 4 y 2 − 2 xy − x 2 + 11 = 0 . Find an equation of the normal to the curve at the point P ( − 3, − 1 ) .
Sometimes functions are given not in the form y = f(x) but in a more complicated form in which it is difficult or impossible to express y explicitly in terms of x. Such functions are called implicit functions. In this unit we explain how these can be differentiated using implicit differentiation.
21 Δεκ 2020 · Figure 2.19:A graph of the implicit function \(\sin (y)+y^3=6-x^2\). Implicit differentiation is a technique based on the Chain Rule that is used to find a derivative when the relationship between the variables is given implicitly rather than explicitly (solved for one variable in terms of the other). We begin by reviewing the Chain Rule.
2y −20 y2 − 16 x2 = dx2 25 y3. dy 6 xy − 4 y2 dy 6 x − 4 y dy 3. Strategy 1: = , Strategy 2: = = dx 3 x2 , Strategy 3: To show all dx 4 x dx 4. 3 x. answers are the same, plug y = into results for strategies 1 and 2. 4.
Implicit Differentiation Practice For each problem, use implicit differentiation to find dy dx in terms of x and y. 1) 2x2 − 5y3 = 2 2) −4y3 + 4 = 3x3 3) 4y2 + 3 = 3x3 4) 5x = 4y3 + 3 5) 2x3 + 5y2 + 2y3 = 5 6) x2 + 5y = −4y3 + 5 7) x + y3 + 2y = 4 8) 2x + 4y2 + 3y3 = 5 9) −5x3y + 2 = x + 2xy2 10) −3x3y2 + 5 = 5x + x2y3
7. implicit derivative dx , x 3 + y 3 = 4. dy. 8. implicit derivative , y 2( y dx 2 − 4 ) = − 5 ) dy. 9. implicit derivative , y = sin ( xy ) dx. dy 2 implicit derivative x ( x y ) 2 2 2. 10. dx , − = x − y. x 2( x 2. Answers.
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