Tough calculus question. calculate dy/dx if Ln(x+y)=e^(x/y)?

Question

I'm having a lot of trouble with this. i get 1/(x+y)(1+dy/dx)=
I don't get how to differentiate the right side. In an example in my book they used 2^2/x and set the exponent to u making it u=2x^-1 and found du/dx= -2x^-2 or -2/x^2. and then plugged into the original problem you get -2e^(2/x)/x^2. i put the paranthesis there so it doesn't get confusing. so if anyone can help me get the right side im sure i'll be able to solve this problem. thanks again =D

my detail button doesn't work so i hadda to post again!! thanks loll. i used what all of you said and i have a multiple choice practice test n im like thinkin why none of the choices have x^2 loll its Ln(x+y)=e^(x/y) is the CORRECT problem geez yall im so sorry lol.

Answer 1

If

ln(x + y) = e^(x/y), then, using the quotient rule on the right hand side in the exponent,

1/(x + y) [1 + dy/dx] = e^(x/y) [ [(1)(y) - x(dy/dx)]/[y^2] ]

1/(x + y) [1 + dy/dx] = e^(x/y) [ [y - x(dy/dx)]/(y^2) ]

1/(x + y) [1 + dy/dx] = e^(x/y) [ 1/y - (x/y^2)(dy/dx) ]

Expanding the left hand side and right hand side,

1/(x + y) + (dy/dx) [1/(x + y)] = [e^(x/y)]/y - [e^(x/y)](x/y^2)(dy/dx)

Moving everything with a dy/dx to the left hand side and everything else to the right hand side,

(dy/dx) [1/(x + y)] + [e^(x/y)](x/y^2)(dy/dx) = [e^(x/y)]/y - 1/(x + y)

dy/dx [1/(x + y) + [e^(x/y)](x/y^2)] = [e^(x/y)]/y - 1/(x + y)

We can replace e^(x/y) with ln(x + y).

dy/dx [1/(x + y) + [ln(x + y)](x/y^2)] = [ln(x + y)]/y - 1/(x + y)

Multiplying both sides by (x + y)(y^2),

dy/dx [y^2 + (x + y)ln(x + y)] = y(x + y)ln(x + y) - y^2

dy/dx = [y(x + y)ln(x + y) - y^2] / [y^2 + (x + y)ln(x + y)]

Answer 2

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Answer 3

I checked all the responses to your previous post of the problem. Most responses there are right even through they have little bit different final forms. Read carefully all those responses!

For your specific question,

d[e^(y/x)]/dx =[d[e^(y/x)]/d(y/x)][d(y/x)/dx] = [e^(y/x)](y'x-y)/x^2

Answer 4

This is not tough
take dreifvative of LHS it is

1/(x+y) d/dx(x+y) = (1+dy/dx)/(x+y)\


take derivative of RHS

e*(x/y) d/dx(x/y)

now you should be able to proceed