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
Derivative of natural log of x+1

Relevant for

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

See proofs

CALCULUS
Derivative of natural log of x+1

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

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

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

Step 5: Final Answer.

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


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

In the instance of this function

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

the graph is illustrated as

graph-of-ln of x+1

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

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

which is illustrated graphically as

graph-of-the derivative of-ln of x+1

Illustrating both graphs in one, we have

graph-of-ln of x+1 and its derivative

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.

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

The final answer is:

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

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.

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

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

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

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

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


See also

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

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