# 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’ll look at how to get the derivative of ln(x+1). We will review some principles, definitions, formulas, graph comparisons of underived and derived ln(x+1), proofs, derivation techniques, and examples.

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

$$y’ = \frac{1}{x+1}$$

## Other Methods to derive the Natural Logarithm of x+1

Aside from the two proofs mentioned above, which also serve as the primary methods for deriving ln(x+1), another method can be utilized to derive this function.

### Derivative of the natural logarithmic of x+1 by using the derivative of general logarithmic function.

Step 1: Express the function as $$f(x) = \log_{e}{(x+1)}$$ instead of $$\ln{(x+1)}$$

Step 2: Take note that the derivative formula for a general logarithmic function is

$$\frac{d}{dx} (\log_{b}{u}) = \frac{1}{u\ln{(b)}} \cdot \frac{d}{dx} (u)$$

where b is any positive real number except 0 and 1 and u is any function other than x which is the argument of the logarithmic function.

Step 3: Get the derivative of $$f(x) = \log_{e}{(x+1)}$$ by applying the formula in step 2.

$$\frac{d}{dx} (\log_{e}{u}) = \frac{1}{u\ln{(e)}} \cdot \frac{d}{dx} (u)$$

In this case, our u is x+1. Hence,

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

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

Step 4: Algebraically evaluate and simplify.

Evaluating ln(e), it is just equal to one. Hence,

$$\frac{d}{dx} (\log_{e}{(x+1)}) = \frac{1}{x+1 \cdot (1)} \cdot (1)$$

Simplifying, we have

$$\frac{d}{dx} (\log_{e}{(x+1)}) = \frac{1}{x+1} \cdot (1)$$

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

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

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

In the instance of this function

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

the graph is illustrated as

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 $$y | y \neq 0$$

## Examples

The examples below show how to derive a natural logarithmic of a variable added to one using the chain rule or the general logarithmic derivative.

### EXAMPLE 1

Derive: $$F(\beta) = \ln{(\beta+1)}$$

Solution: Using chain rule, we have

Step 1: Analyze if the natural logarithm of $$\beta+1$$ is a function of the same variable $$\beta$$. In this problem, it is. Hence, proceed to step 2.

Step 2: Isolate the two functions composed in $$F(\beta)$$. The outer function is

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

and the inner function is

$$u = g(\beta) = \beta+1$$

Step 3: Derive both outer and inner functions separately using their respective base derivative formula.

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

$$g'(\beta) = \frac{d}{d\beta} (\beta+1) = 1$$

Step 4: Multiply the derivative of $$f(u)$$ and $$g(\beta)$$.

$$F'(\beta) = \left( \frac{1}{u} \right) \cdot (1)$$

Step 5: Substitute u into the derivative of $$f(u)$$.

$$F'(\beta) = \left( \frac{1}{(\beta+1)} \right) \cdot (1)$$

Step 6: Simplify and declare the final answer.

$$F'(\beta) = \frac{1}{(\beta+1)}$$

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

### EXAMPLE 2

Derive: $$f(\lambda) = \ln{(\lambda+1)}$$

Solution: Using the general logarithmic derivative formula, we have

Step 1: Express the function as $$f(\lambda) = \log_{e}{(\lambda+1)}$$ instead of $$\ln{(\lambda+1)}$$

Step 2: Set the logarithmic argument as u.

$$u = \lambda + 1$$

Step 3: Recalling the derivative formula for a general logarithmic function, we have

$$\frac{d}{dx} (\log_{b}{u}) = \frac{1}{u\ln{(b)}} \cdot \frac{d}{dx} (u)$$

Step 4: Derive $$f(\lambda) = \log_{e}{(\lambda+1)}$$ by applying the formula in step 2.

$$\frac{d}{d\lambda} (\log_{e}{(\lambda+1)}) = \frac{1}{u\ln{(e)}} \cdot \frac{d}{d\lambda} (u)$$

$$\frac{d}{d\lambda} (\log_{e}{(\lambda+1)}) = \frac{1}{(\lambda+1)\ln{(e)}} \cdot \frac{d}{d\lambda} (\lambda+1)$$

$$\frac{d}{d\lambda} (\log_{e}{(\lambda+1)}) = \frac{1}{(\lambda+1) \cdot (1)} \cdot (1)$$

$$\frac{d}{d\lambda} (\log_{e}{(\lambda+1)}) = \frac{1}{\lambda+1}$$

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