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Simple Implicit Differentiation Problem

Problem Statement
Implicitly differentiate ##ln(x+y)=x## and solve for ##\frac{dy}{dx}## in terms of ##x##
Relevant Equations
How to differentiate the natural logarithm; ##\frac{d(ln(f(x)+g(x))}{dx} = \frac{f'(x)+g'(x)}{f(x)+g(x)}##
Okay so I'm really not sure where I went wrong here; here's how I worked through it:

$$\ln\left(y+x\right)=x$$
$$\frac{\frac{dy}{dx}+1}{y+x}=1$$
$$\frac{dy}{dx}+1=y+x$$

If ##\ln\left(y+x\right)=x## then ##y+x=e^x## and ##y=e^x-x##

$$\frac{dy}{dx}=y+x-1$$
$$\frac{dy}{dx}=e^x-x+x-1$$
$$\frac{dy}{dx}=e^x-1$$

So my answer is ##e^x-1##, however the answer in the back of the book says it should be ##e-1##. Can anyone give me an idea of where I've gone wrong with this question?
 

Orodruin

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##y## clearly cannot be a constant. Are you sure the problem is not asking for ##y'(1)##, which based on your findings is ##y'(1) = e^1-1 = e-1##? Alternatively, the ##x## has just fallen away as a typo in the answers.
 

BvU

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Printing error in the back of the book
 
##y## clearly cannot be a constant. Are you sure the problem is not asking for ##y'(1)##, which based on your findings is ##y'(1) = e^1-1 = e-1##? Alternatively, the ##x## has just fallen away as a typo in the answers.
After going onto the next question I found that it does ask to find ##y'(1)##, however the back of the book says that ##e-1## is the answer for both this question and the previous question, so it must have just been a printing error.
 
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Edit: Never mind.
I misread the previous post and was thinking the OP was trying to solve the differential equation.

Aside from the implicit differentiation part, the problem is one about a differential equation. Was there an initial condition given?
Without an initial condition, the solutions are an entire family of functions. In this case, your third equation in post #1 is ##y' - y = x - 1##
The general solution of this equation is ##y = Ce^x - x##, where the constant C can be determined only if we have an initial condition, i.e., a given point on the graph of the solution function. Since the problem mentions y'(1), I suspect that the value of y(1) is given somewhere.
 
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Orodruin

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It is hardly about a differential equation when you are given ##\ln(y+x) = x##, which is not a differential equation. You can solve for ##y## directly from there.
 
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It is hardly about a differential equation when you are given ##\ln(y+x) = x##, which is not a differential equation. You can solve for ##y## directly from there.
I agree. Apparently @Saracen Rue went beyond the problem asked for, part of which was "and solve for ##\frac{dy}{dx}## in terms of ##x##".

My response was in reply to his work in solving the DE.
 

Orodruin

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I agree. Apparently @Saracen Rue went beyond the problem asked for, part of which was "and solve for ##\frac{dy}{dx}## in terms of ##x##".

My response was in reply to his work in solving the DE.
I disagree, he did not solve the differential equation. He solved ##\ln(y+x) = x## and he computed ##dy/dx## in terms of ##x##.
 
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I disagree, he did not solve the differential equation. He solved ##\ln(y+x) = x## and he computed ##dy/dx## in terms of ##x##.
OK, I didn't read his post carefully , mistakenly thinking he had attempted to solve a DE.
 

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