Simple Implicit Differentiation Problem

In summary, Saracen Rue found that if the equation is written in the form ##\ln(y+x) = x## then ##y+x=e^x## and ##y=e^x-x## and so ##e^x-1## is the solution.
  • #1
Saracen Rue
150
10
Homework 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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  • #2
##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.
 
  • #3
Printing error in the back of the book
 
  • #4
Orodruin said:
##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.
 
  • #5
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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  • #6
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.
 
  • #7
Orodruin said:
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.
 
  • #8
Mark44 said:
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##.
 
  • #9
Orodruin said:
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.
 

What is simple implicit differentiation?

Simple implicit differentiation is a method used in calculus to find the derivative of a function that is not explicitly written in terms of one variable. It involves using the chain rule and implicit differentiation to find the derivative.

Why is simple implicit differentiation useful?

Simple implicit differentiation is useful because it allows us to find the derivative of functions that cannot be easily solved using traditional methods. It is also useful in solving optimization problems and finding tangent lines to curves.

How do you perform simple implicit differentiation?

To perform simple implicit differentiation, you first need to identify the dependent and independent variables in the function. Then, use the chain rule to find the derivative of the dependent variable with respect to the independent variable. Finally, solve for the derivative using algebraic manipulation.

What are some common mistakes when using simple implicit differentiation?

Some common mistakes when using simple implicit differentiation include forgetting to use the chain rule, incorrectly identifying the dependent and independent variables, and making algebraic errors when solving for the derivative.

Can simple implicit differentiation be used for all functions?

No, simple implicit differentiation can only be used for functions that are differentiable. This means that the function must be continuous and have a defined derivative at every point. It also cannot be used for functions with multiple variables or functions that are not functions of x.

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