The natural logarithm of x+1, also denoted as ln(x+1), is the logarithm of *x*+1 to base *e (euler’s number)*. **The derivative of the natural logarithm of x+1 is equal to one over x+1, 1/(x+1)**. This derivative can be found using the chain rule or with implicit differentiation.

In this article, we will learn how to obtain the derivative of ln(x+1). We will learn about proofs and graphical comparisons of ln(x+1) and its derivative.

##### CALCULUS

**Relevant for**…

Learning how to find the derivative of the natural logarithm of x+1.

##### CALCULUS

**Relevant for**…

Learning how to find the derivative of the natural logarithm of x+1.

## Proofs of the Derivative of Natural Logarithm of *x+1*

Listed below are the proofs of the derivative of * \(\ln{(x+1)}\)*. These proofs can also serve as the main methods of deriving this function.

### Proof of the derivative of *ln(x+1)* using the Chain Rule Formula

In the derivative process of the natural log of *x+1*, the chain rule formula is used to verify the derivative formula for the natural log of *x*+1 since it is made up of these two functions.

The natural logarithmic function will be the outer function *f(u)* in the composite function *ln(x+1)*, whereas the binomial *x+1* will be the inner function *g(x)*.

You can review the chain rule formula by looking at this article: Chain Rule of derivatives. You can also look at this article for the proof of the natural logarithm’s derivative using limits: Derivative of Natural log (ln(x)).

Let’s have the derivative of the function

$$ F(x) = \ln{(x+1)}$$

We can figure out the two functions that make up *F(x)*. There is a natural logarithmic function and a monomial in this case. The outer function can be configured as follows.

$$ f(u) = \ln{(u)}$$

where

$$ u = x+1$$

Setting the binomial *x+1* as the inner function of *f(u)* by denoting it as *g(x)*, we have

$$ f(u) = f(g(x))$$

$$ g(x) = x+1$$

$$ u = g(x)$$

Deriving the outer function *f(u)* using the derivative of natural log in terms of *u*, we have

$$ f(u) = \ln{(u)}$$

$$ f'(u) = \frac{1}{u}$$

Deriving the inner function *g(x)* using power rule since it is a monomial, we have

$$ g(x) = x+1$$

$$ g'(x) = 1$$

Algebraically multiplying the derivative of outer function $latex f'(u)$ by the derivative of inner function $latex g'(x)$, we have

$$ \frac{dy}{dx} = f'(u) \cdot g'(x)$$

$$ \frac{dy}{dx} = \left(\frac{1}{u} \right) \cdot (1)$$

Substituting *u* into *f'(u)*, we have

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

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

As a result, we arrive at the *ln(x+1)* derivative formula.

$$ \frac{d}{dx} \ln{(x+1)} = \frac{1}{x+1}$$

### Proof of the derivative of *ln(x+1)* using implicit differentiation

You are advised to learn/review the derivatives of exponential functions and implicit differentiation for this proof.

Given that the equation

$$ y = \ln{(x+1)}$$

In general logarithmic form, it is

$$ \log_{e}{(x+1)} = y$$

And in exponential form, it is

$$ e^y = x+1$$

Implicitly deriving the exponential form in terms of *x*, we have

$$ e^y = x+1$$

$$ \frac{d}{dx} (e^y) = \frac{d}{dx} (x+1) $$

$$ e^y \cdot \frac{dy}{dx} = 1 $$

Isolating \( \frac{dy}{dx} \), we have

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

We recall that in the beginning, \( y = \ln{(x+1)} \). Substituting this to the *y* of our derivative, we have

$$ \frac{dy}{dx} = \frac{1}{e^{(\ln{(x+1)})}} $$

Simplifying and applying a property of logarithm, we have

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

Evaluating, we now have the derivative of \( y = \ln{(x+1)} \)

$$ y’ = \frac{1}{x+1} $$

## Graph of *ln(x+1)* vs. its derivative

We have the function

$$ f(x) = \ln{(x+1)}$$

and its graph is

And as we learned above, deriving \(f(x) = \ln{(x+1)}\) will be

$$ f'(x) = \frac{1}{x+1}$$

which is illustrated graphically as

Illustrating both graphs in one, we have

By examining the differences between these functions using these graphs, you can see that the original function \(f(x) = \ln{(x+1)}\) has a domain of

\( (-1,\infty) \) or \( x | x > -1 \)

and lies within the range of

\( (-\infty, \infty) \) or *all real numbers*

whereas the derivative \(f'(x) = \frac{1}{x+1}\) has a domain of

\( (-\infty,-1) \cup (-1,\infty) \) or \( x | x \neq -1 \)

which lies within the range of

\( (-\infty,0) \cup (0,\infty) \) or \( y | y \neq 0 \)

## See also

Interested in learning more about the derivatives of logarithmic functions? Take a look at these pages:

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