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anonymous
 5 years ago
Find dy/dx by implicit differentiation.
e^(x / y) = 9x  y
anonymous
 5 years ago
Find dy/dx by implicit differentiation. e^(x / y) = 9x  y

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Dx(e^(x/y)) = e^(x/y) * (yxy')/(y^2) Dx(9xy) = 9 y' e^(x/y) * (yxy')/(y^2) = 9 y' e^(x/y) * [(1/y)xy'/y^2] = 9  y' [e^(x/y)]/y  e^(x/y)xy'/y^2 = 9 y' [e^(x/y)]/y 9 = e^(x/y)xy'/y^2 y' [e^(x/y)]/y 9 = y'[e^(x/y)x/y^2 1] [e^(x/y)]/(y 9)(e^(x/y)x/y^2 1) = y' As long as I didnt get lost in that, it should be it :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0to clean it up some: lets say a=e^(x/y) then, a/[y9][ax/(y^2) 1] = y'

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I did get lost.... so lets try that again :) a(yxy')/y^2 = 9  y' ...... Dx(LHS) = Dx(RHS); a=e^(x/y) (ay/y^2)  (axy'/y^2) = 9  y' .....distrubute a thru LHS (a/y)  9 = (ax/y^2)(y')  (y') .....()9; (+)axy'/y^2; and simplify ay/y^2 [(a9y)/y] = y'[ (ax/y^2)  1] .... undistribute (y') from RHS; simplify LHS [(a9y)/y] / [(ax  y^2)/y^2] = y' .....simplify [(ax/y^2)  1] and (divide) [ y(a9y) ]/[ax  y^2] = y' ...... simplify LHS Last step is to replace a with e^(x/y).
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