# Derivative of ln(x+1) with Proofs and Graphs

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.

See proofs

##### CALCULUS

Relevant for

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

See proofs

## 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 $$y | y \neq 0$$