# Simple Implicit Differentiation Problem

#### Saracen Rue

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?

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#### 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

Homework Helper
Printing error in the back of the book

#### Saracen Rue

$y$ 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.

#### Mark44

Mentor
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.

Last edited:

#### 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.

#### Mark44

Mentor
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 $x$.

#### Mark44

Mentor
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.

"Simple Implicit Differentiation Problem"

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