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- 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?

$$\ln\left(y+x\right)=x$$

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

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

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?