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:
Jefferson Huera Guzman
Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.
