Implicit Differentiation Answers

CALCULUS 1-NOTES Name SECTION 3.6 (DAY 1) Date Block 3.6 Implicit Differentiation (Day 1) Implicit differentiation is nothing more than a special case of the chain rule for derivatives. It is used to solve equations such as x ^ 3 - y ^ 3 + 7x = 0 which does not let us solve for y in terms of x easily. The majority of differentiation problems in first-year calculus involve functions y written EXPLICITLY as functions of x. For example, if then, the derivatives are y = 3x ^ 2 - sin(7x + 5) y' = 6x - 7cos(7x + 5) OR y = 3x ^ 2 + 4x y' = 6x + 4 However, some functions y are written IMPLICITLY as functions of x. An example of this is the equation: x ^ 2 + y ^ 2 = 25 which represents a circle of radius five centered at the origin. Suppose that we wish to find the slope of the line tangent to the graph of this equation at the point (3,-4). x ^ 2 + y ^ 2 - 25 (3,-4) How could we find the derivative of y in this instance? One way is to first write y explicitly as a function of x. (solve for y=) x ^ 2 + y ^ 2 = 25 y ^ 2 = 25 - x ^ 2 y = plus/minus sqrt(25 - x ^ 2) y = - sqrt(25 - x ^ 2) = - (25 - x ^ 2) ^ (1/2) CR y' = - 1/2 * (25 - x ^ 2) ^ (- 1/2) * (- 2x) = x/(sqrt(25 - x ^ 2)) X The derivative of y is: Thus, the slope of the line tangent to the graph at the point (3,-4) is m = 3/(sqrt(25 * 9)) = 32/(sqrt(16)) = 3/4 Note: the positive square root represents the top semi-circle and the negative square root represents the bottom semi-circle. Unfortunately, not every equation involving x and y can be solved explicitly for y. So let's do the same problem implicitly. Take the derivate of both sides w/respect to x Because it's not an x use the chain rule x ^ 2 + y ^ 2 = 25 d/dx (x ^ 2 + y ^ 2) = d/dx (25) d/dx (x ^ 2) + d/dx (y ^ 2) = d/dx (25) 2x + 2y * d/dx (y) = 0 2y * d/dx (y) = - 2x dy=-2x-x ax 2y Y Thus, the slope of the line tangent to the graph at the point (3, 4) is m = - 3/- 4 = 3/4 -same slope as when solved explicitly The second method illustrates the process of implicit differentiation. It is important to note that the derivative expression for explicit differentiation involves x only, while the derivative expression for implicit differentiation may involve both x and y. Steps for Implicit Differentiation 1) Differentiate both sides of the equation with respect to x. 2) Collect the terms with d/dx (y) on one side of the equation 3) Factor out d/dx (y) when there is more than one y) 4) Solve for d/du (y) by dividing. kanytime you take the derivative of y_{2} you attach d/dx (y) Examples: Find d/dx (y) of the each of the following using implicit differentiation. (Solve for ax) (1) 3x ^ 3 - 4y ^ 2 = 2x 9x ^ 2 - 8y * d/dx (y) = 2 - 8y * d/dx (y) = 2 - 9x ^ 2 d/dx (y) = (2 - 9x ^ 2)/(- 8y) d/dx (y) = (- 2 + 9x ^ 2)/(8y) 2) x = tan y I = sec^2 y * d/dx (y) 1/(sec^2 y) = d/dx (y) cos^2 y = d/dx (y) PR 3) 3x ^ 2 * y + 2x =4y+ 6x - y ^ 9 + 3x ^ 2 * d/dx (y) + 2 = 12x ^ 2 6xy + 3x ^ 2 * d/dx (y) = 12x ^ 2 - 2 3x ^ 2 * d/dx (y) = 12x ^ 2 - 6xy - 2 d/dx (y) = (12x ^ 2 - 6xy - 2)/(3x ^ 2) X y ^ 2 = (1 - x ^ 2)/(x ^ 2) * QR * R ^ 2 2y * d/dx (y) = (x ^ 2 - (- 3x ^ 2) - (i - x ^ 3) * 2x)/((x ^ 2) ^ 2) 2y * d/dx (y) = (- 3x ^ 4 - 2x + 2x ^ 4)/(x ^ 4) 2y d/dx (y) = (- x ^ 4 - 2x)/(x ^ 4) dY d x = - X ^ 4 - 2x X^ 4 X^ 4 1 . 1 7 d/dx (y) = (- x ^ 4 - 2x)/(2y * x ^ 4) CALCULUS I-NOTES Name SECTION 3.6 (DAY 2) Date Block 3.6 Implicit Differentiation (Day 2) Examples: Find the derivative of each of the following y ^ 3 + 7y = x ^ 3 3y ^ 2 * d/dx (y) + 7 * d/dx (y) = 3x ^ 2 d/dx (y) * (3y ^ 2 + 7) = 3x ^ 2 (d/dx (y)) / ((3x ^ 2)/(3y ^ 2 + 7)) 4x ^ 2 * y - 3y = x ^ 3 - 1 4x ^ 2 * d/dx (y) + y8x - 3 * d/dx (y) = 3x ^ 2 4x * 2/(dx) * 3 * d/dx (y) = 3x ^ 2 - 8xy d/dx (y) * (4x ^ 2 - 3) = 3x ^ 2 - 8xy d/dx (y) = (3x ^ 2 - 8xy)/(4x ^ 2 - 3) 5y y ^ 4 + 3xy = 2x ^ 3 + 1 4y ^ 3 * d/dx (y) + 3x * d/dx (y) + 3y = 6x ^ 2 * y d/dx (y) * (4y ^ 3 + 3x) = 6x ^ 2 - 3y d/dx (y) = (b * x ^ 2 - 3y)/(4y ^ 3 + 3x) (1E)/(y ^ 2) * y ^ 3 + 3y ^ 7 - 5x ^ 2 = 7 +2x3+2lydy-10x = 0 2 dx d/dx (y) * (3x ^ 2 * y ^ 2 + 21y ^ 6) = - 2x * y ^ 3 + 10x d/dx (y) = (- 2x * y ^ 3 + i0x)/(3x ^ 2 * y ^ 2 + 2i * y ^ 6) CR 7) y ^ 3 - x * y ^ 2 + cos(xy) = 2 3y ^ 2 * d/dx (y) - (1y ^ 2 + x * 2y * d/dx (y)) - sin(xy) * (1y + x * d/dx (y)) = 0 disin bute the naye reorder distribulk -sinlay) (reorder 3y²²-2xy-ysin(xy)-xsin(xy)=0 2 + y sin (xy) move non- terms to the other side +y² +sin(xy) 3y ^ 2 * (dy)/(dx) - 2xy * (dy)/(dx) - x * (dy)/(dx) * sin(xy) =y^ 2 +y sin)(xy) take ax + d/dx (y) dy dx ( 3y ^ 2 - 2xy - x * sin(xy) )=y^ 2 +y sin(xy) 3y^ 2 -2xy-x sin(xy) DIVIDE d/dx (y) = (y ^ 2 + y * sin(xy))/(3y ^ 2 - 2xy - x * sin(xy))