Question

Find dy/dx by implicit differentiation.
e^(x / y) = 9x - y

Answer 1

Dx(e^(x/y)) = e^(x/y) * (y-xy')/(y^2)
Dx(9x-y) = 9 -y'

e^(x/y) * (y-xy')/(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 :)

Answer 2

to clean it up some: lets say a=e^(x/y)

then, a/[y-9][ax/(y^2) -1] = y'

Answer 3

I did get lost.... so lets try that again :)

a(y-xy')/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

[(a-9y)/y] = y'[ (ax/y^2) - 1] .... undistribute (y') from RHS; simplify LHS

[(a-9y)/y] / [(ax - y^2)/y^2] = y' .....simplify [(ax/y^2) - 1] and (divide)

[ y(a-9y) ]/[ax - y^2] = y' ...... simplify LHS

Last step is to replace a with e^(x/y).

Answer 4

thank u