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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.
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 ) .
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
Implicit di erentiation is a method for nding the slope of a curve, when the equation of the curve is not given in \explicit" form y = f(x), but in \implicit" form by an equation g(x;y) = 0.
Implicit Differentiation Example: 7Find × ì × ë for 𝑦 6 F5𝑥 L3𝑦 Step 1: Take the derivative. Each time the derivative of “y” is involved, include a × ì × ë. Step 2: Gather all terms with × ì × ë on the left side, everything else on the right. Step 3: Factor out the × ì × ë if necessary, to create only one × ì × ë
differentiation. Example A: Given the equation 535x2 + 2 y 2 = , a) Verify that the point ( x , y ) = (– 3, 2) satisfies the equation. b) Use implicit differentiation to find
Example 1: Find if. Solution: Differentiating both sides with respect to and then solving for. Solving for we finally obtain . Implicit differentiation can be used to calculate the slope of the tangent line as the example below shows. Example 2: Find the equation of the tangent line that passes through point to the graph of. Solution:
