#$&* course Mth 271 11/10 830p Query problem 7th edition 2.5.48 2.5.44 der of 3/(x^3-4)^2 **** What is your result?
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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. DER** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): n/a ------------------------------------------------ Self-critique Rating:3 ********************************************* 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: f(x) = 1/u^½ f’(x) = 2x-3
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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): I see that 2x-3 comes from (x^2-3x+4), but I thought that was to the -½ so it should have been (½)(x) - 3. I used it because it was used in the solution but I am not sure about it. ------------------------------------------------ Self-critique Rating:2 ********************************************* 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: f(x) = .25z^(½) f’(x) = .25((½)z^(-½) g(x) = (.5n^2+5n+25) g’(x) = 2(.5)n+5 = n+5 (1/(8* sq. rt. of z)) * (n+5) (n+5)/(8* sq. rt. of .5n^2+5n+25) = 17/(8*sq. rt. of .5*12^2 + 5(12) +25) = 17/(8* sq. rt. of 157)= 17/100.24 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): n/a ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!