#$&* course Mth 271 ** Query problem 7th edition 2.5.48 2.5.44 der of 3/(x^3-4)^2 **** What is your result?
.............................................
Given Solution: `a This function can be expressed as f(g(x)) for g(x) = x^3-4 and f(z) = 3 / z^2. The 'inner' function is x^3 - 4, the 'outer' function is 3 / z^2 = 3 z^(-2). So f ' (z) = -6 / z^3 and g'(x) = 3x^2. Thus f ' (g(x)) = -6/(x^3-4)^3 so the derivative of the whole function is [3 / (x^3 - 4) ] ' = g ' (x) * f ' (g(x)) = 3x^2 * (-6/(x^3-4)^3) = -18 x^2 / (x^3 - 4)^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q **** Query problem 2.5.62 tan line to 1/`sqrt(x^2-3x+4) at (3,1/2) **** What is the equation of the tangent line? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: d/dx [1/(sqrt(x^2 - 3x + 4)] = 1/(x^2 - 3x + 4) ^ 1/2 = (x^2 - 3x + 4) ^ -1/2 = -1/2 (x^2 - 3x + 4) ^ -3/2 * ( 2x - 3) = -1/2 (2x - 3) * (x^2 -3x + 4) ^ -3/2 X = 3 d/dx = - 3/2 * 4^ -3/2 = -3/16 Equation of tangent line: (y - ½) = -3/16 (x - 3) / * 16 16y - 8 = -3 (x-3) 16y - 8 = -3x + 9 16y = - 3x + 17 Y = -3/16 x + 17/16 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a The derivative is (2x - 3) * -1/2 * (x^2 - 3x + 4) ^(-3/2) . At (3, 1/2) we get -1/2 (2*3-3)(3^2- 3*3 + 4)^(-3/2) = -1/2 * 3 (4)^-(3/2) = -3/16. The equation is thus ( y - 1/2) = -3/16 * (x - 3), or y = -3/16 x + 9/16 + 1/2, or y = -3/16 x + 17/16. DER** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q **** Query problem 2.5.68 rate of change of pollution P = .25 `sqrt(.5n^2+5n+25) when pop n in thousands is 12 **** At what rate is the pollution changing at the given population level? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y’ = 0.25 * (0.5n^2 + 5n + 25) ^ ½ = 0.25 * [1/2 (0.5n^2 + 5n + 25) ^ -1/2 * (n+ 5)] Y’ = 0.25 * (0.5 (n+5))/(sqrt(0.5n^2 + 5n + 25)) Y’ = (0.125 (n+5))/(sqrt(0.5n^2 + 5n + 25) n = 12 y’ = 0.17 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a The derivative is .25 [ (n + 5) * 1/2 * (.5 n^2 + 5 n + 25) ^(-1/2) ) = (n+5) / [ 8 `sqrt(.5n^2 + 5n + 25) ] When n = 12 we get (12+5) / ( 8 `sqrt(.5*12^2 + 5 * 12 + 25) ) = 17 / 100 = .17, approx. DER** ADDITIONAL COMMENT Details of calculating P ': P is of the form f(g(x)) with g(x) = .5 n^2 + 5 n + 25 and f(z) = .25 z^(1/2). g ' (x) = n + 5 and f ' (z) = .25 ( 1/2 z^(-1/2) ) = 1 / (8 z^(1/2)), or -1 / (8 sqrt(z) ). Thus P ' = g ' (x) * f ' (g(x)) = (n + 5) * (1/ (8 sqrt(n^2 + 5 n + 25) ) ) = (n+5) / (8 sqrt(n^2 + 5 n + 25). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!