Differentiation of implicit functions LO To differentiate implicit

Differentiation of implicit functions LO: To differentiate implicit functions. www. mathssupport. org

Implicit functions The functions we have studied so far have been explicitly defined, the dependent variable is defined in terms of the independent variable. Relations where is often difficult or impossible to make y the subject of the formula are called implicit relations From the list, identify the implicit relations. 2 x + 3 y = 8 x + x 2 y + y 3 = 100 x 3 2 xy = 5 5 x 2 + y = 2 2 x 2 + y 3 = 1 www. mathssupport. org

Implicit differentiation Example 1: Differentiate y 2 = 4 x implicitly with respect to x Differentiate the left-hand side Differentiate the right-hand side with respect to x using the chain rule. with respect to x = 4 = = 2 y Hence: 2 y = 4 = Therefore: www. mathssupport. org 4 2 y 2 = y

Implicit differentiation Example 2: Differentiate x 2 + y 2 = 4 implicitly with respect to x We simply differentiate term by term across the equation. x 2 + y 2 = 4 + = 2 x + Hence: = 0 2 x + 2 y 2 y Therefore: www. mathssupport. org = 0 = 2 x = 2 x x = y 2 y Note that using the chain rule n– 1 = ny

Implicit differentiation Example 3: Differentiate x 2 + y 3 = 8 implicitly with respect to x We differentiate term by term across the equation. x 2 + y 3 = 8 + Hence: = 2 x + 3 y 2 = 0 = 2 x 2 x = 2 3 y www. mathssupport. org n– 1 = ny

Implicit differentiation Example 4: Differentiate x + x 2 y + y 3 = 100 implicitly with respect to x We differentiate term by term across the equation. x + x 2 y + y 3 = 100 + + 1 + x 2 Rearranging Factorising + 2 x y + 3 y 2 Using the product rule = u = x 2 = 0 + 3 y 2 = 1 2 xy (x 2 + 3 y 2) = 1 2 xy x 2 Rearranging www. mathssupport. org = 1 2 xy x 2 + 3 y 2 v = y n– 1 = ny = = 2 x = x 2 + 2 x y

Implicit differentiation Example 5: Differentiate x 3 + y 3 9 xy = 0 implicitly with respect to x We differentiate term by term across the equation. x 3 + y 3 9 xy = 0 + 3 x 2 + 3 y 2 Rearranging Factorising 9 x 3 y 2 (3 y 2 9 x) Rearranging www. mathssupport. org = 9 y = 0 Using the rule product rule v = y u = 9 x n– 1 = ny = =9 = 9 y 3 x 2 2 = 9 y 3 x 2 3 y x 2 = = 2 2 3 y 9 x y 3 x = 9 x + 9 y

Implicit differentiation Example 6: The point P(2, m), where m < 0, lies on the curve 2 x 2 y + 3 y 2 = 16 (a) Calculate the value of m. Substitute (2, m) into the implicit function for (x, y). 2 x 2 y + 3 y 2 = 16 2(2)2(m) + 3(m)2 = 16 Ordering the quadratic equation 8 m + 3 m 2 = 16 3 m 2 + 8 m +16 = 0 Factorising (3 m 4) (m + 4) = 0 Solving for m Since m < 0, www. mathssupport. org 3 m 4 = 0 or m + 4 = 0 or m = 4

Implicit differentiation Example 6: The point P(2, m), where m < 0, lies on the curve 2 x 2 y + 3 y 2 = 16 (b) Find the gradient of the normal to the tangent at P. Differentiate implicitly 2 x 2 y + 3 y 2 = 16 = + 2 x 2 Rearranging Factorising + 4 x y + 6 y 2 x 2 (2 x 2 Rearranging www. mathssupport. org + 6 y) =0 = 4 xy = 2 2 x + 6 y At P(2, -4) rule Using the product rule 4(2)(-4) 2 2 v = y u == 2 x 2(2) + 6(-4) n– 1 = ny 2 = ==4 x = 2 x 2 + 4 x y Hence, the gradient of the normal at P is:

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