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6 Nov 2019
y = [(3x-4)5 / (x2 + 2)3]1 / 2 ln y = ln [(3x-4)5 / (x2 + 2)3]1 / 2 Apply log rule: ln (ab) = b ln a ln y = 1 / 2 ln[(3x-4)5 / (x2 + 2)3] Apply log rule: ln a / b = ln a-ln b ln y = 1 / 2 [ln(3x-4)5 -ln (x2 +2 )3] Apply log rule: ln (ab) = b ln a ln y = 1 / 2 [5 ln (3x-4)-3ln (x2 + 2)] ln y = 5 / 2 ln (3x-4)-3 / 2 ln (x2 + 2) d / dx [ln y]= 5 / 2 d / dx [ln (3x-4)]- 3 / 2 d / dx [ln (x2 + 2)] 1 / y dy / dx = 5 / 2 . 1 / 3x-4 . d / dx (3x-4)-3 / 2 . d / dx (x2 + 2) 1 / y d / dx = 5 / 2 . 1 / 3x-4(3)-3 / 2 . 1 / x2 + 2 (2x) 1 / y dy / dx = [15 / 2(3x-4)-3x / x2 + 2 dy / dx = [15 / 2(3x-4) - 3x / x2 + 2] y put y = , we get dy / dx = [15 / 2(3x-4) - 3x / x2 + 2] is the required answer. we multiplied 3x5 to get 15. why not 2x times 3/2? what happens to this 2x? where did it go? Show transcribed image text
y = [(3x-4)5 / (x2 + 2)3]1 / 2 ln y = ln [(3x-4)5 / (x2 + 2)3]1 / 2 Apply log rule: ln (ab) = b ln a ln y = 1 / 2 ln[(3x-4)5 / (x2 + 2)3] Apply log rule: ln a / b = ln a-ln b ln y = 1 / 2 [ln(3x-4)5 -ln (x2 +2 )3] Apply log rule: ln (ab) = b ln a ln y = 1 / 2 [5 ln (3x-4)-3ln (x2 + 2)] ln y = 5 / 2 ln (3x-4)-3 / 2 ln (x2 + 2) d / dx [ln y]= 5 / 2 d / dx [ln (3x-4)]- 3 / 2 d / dx [ln (x2 + 2)] 1 / y dy / dx = 5 / 2 . 1 / 3x-4 . d / dx (3x-4)-3 / 2 . d / dx (x2 + 2) 1 / y d / dx = 5 / 2 . 1 / 3x-4(3)-3 / 2 . 1 / x2 + 2 (2x) 1 / y dy / dx = [15 / 2(3x-4)-3x / x2 + 2 dy / dx = [15 / 2(3x-4) - 3x / x2 + 2] y put y = , we get dy / dx = [15 / 2(3x-4) - 3x / x2 + 2] is the required answer. we multiplied 3x5 to get 15. why not 2x times 3/2? what happens to this 2x? where did it go?
Show transcribed image text Beverley SmithLv2
2 Jul 2019