Derivative of ln(2x) with Proofs and Graphs

The natural logarithm of 2x, also denoted as ln(2x), is the logarithm of 2x to base e (euler’s number). The derivative of the natural log of 2x is equal to one over x, 1/x. We can find this derivative using the chain rule or with implicit differentiation.

In this article, we’ll look at how to compute the derivative of ln(2x). We will go over some fundamentals, definitions, formulas, graph comparisons of underived and derived ln(2x), proofs, derivation techniques, and some examples.

CALCULUS
Derivative of natural log of 2x

Relevant for

Learning how to find the derivative of the natural logarithm of 2x.

See proofs

CALCULUS
Derivative of natural log of 2x

Relevant for

Learning how to find the derivative of the natural logarithm of 2x.

See proofs

Proofs of the Derivative of Natural Logarithmic of 2x

You may wonder why the derivative of \( \ln{(x)} \) and \( \ln{(2x)} \) are just the same. To show why, here are some proofs of the derivative of \( \ln{(2x)} \).

These proofs can also serve as the main methods of deriving 2x.

Proof of the derivative of ln(2x) using the Chain Rule Formula

Given that this is a composed function, the chain rule formula can be used to prove the derivative formula for the natural log of 2x. In the composite function ln(2x), the natural logarithmic function will be the outer function f(u), while the monomial 2x will be the inner function g(x).

As a prerequisite for this topic, please review the chain rule formula by looking at this article: Chain Rule. You may also check out this article for the proof of the derivative of natural logarithm using limits: Derivative of Natural log (ln(x)).

Let’s have the derivative of the function

$$ F(x) = \ln{(2x)}$$

We can determine the two functions that comprise F(x). In this instance, there is a natural logarithmic function and a monomial. We may configure the outer function as follows:

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

where

$$ u = 2x$$

Setting the monomial 2x as the inner function of f(u) by denoting it as g(x), we have

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

$$ g(x) = 2x$$

$$ u = g(x)$$

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

$$ g(x) = 2x$$

$$ g'(x) = 2$$

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

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

$$ \frac{dy}{dx} = \left(\frac{1}{u} \right) \cdot (2)$$

Substituting u into f'(u), we have

$$ \frac{dy}{dx} = \left(\frac{1}{(2x)} \right) \cdot (2)$$

$$ \frac{dy}{dx} = \frac{2}{2x}$$

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

As a result, we arrive at the ln(2x) derivative formula.

$$ \frac{d}{dx} \ln{(2x)} = \frac{1}{x}$$

Proof of the derivative of ln(2x) using implicit differentiation

In this proof, you are hereby recommended to learn/review the derivatives of exponential functions and implicit differentiation.

Suppose we have the equation

$$ y = \ln{(2x)}$$

In general logarithmic form, it is

$$ \log_{e}{2x} = y$$

And in exponential form, it is

$$ e^y = 2x$$

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

$$ e^y = 2x$$

$$ \frac{d}{dx} (e^y) = \frac{d}{dx} (2x) $$

$$ e^y \cdot \frac{dy}{dx} = 2 $$

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

$$ \frac{dy}{dx} = \frac{2}{e^y} $$

We recall that in the beginning, \( y = \ln{(2x)} \). Substituting this to the y of our derivative, we have

$$ \frac{dy}{dx} = \frac{2}{e^{(\ln{(2x)})}} $$

Simplifying and applying a property of logarithm, we have

$$ \frac{dy}{dx} = \frac{2}{2x} $$

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

Evaluating, we now have the derivative of \( y = \ln{(2x)} \)

$$ y’ = \frac{1}{x} $$


Other Methods to derive the Natural Logarithmic of 2x

Aside from the two proofs given above which also serve as the main methods of deriving ln(2x), an additional method can also be used to derive this same function.

Derivative of the natural logarithmic of 2x by using the derivative of general logarithmic function.

Step 1: Express the function as \(f(x) = \log_{e}{2x}\) instead of \(\ln{(2x)}\)

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 \(f(x) = \log_{e}{2x}\) 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 2x. Hence,

$$ \frac{d}{dx} (\log_{e}{2x}) = \frac{1}{2x\ln{(e)}} \cdot \frac{d}{dx} (2x)$$

$$ \frac{d}{dx} (\log_{e}{2x}) = \frac{1}{2x\ln{(e)}} \cdot (2)$$

Step 4: Algebraically evaluate and simplify.

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

$$ \frac{d}{dx} (\log_{e}{2x}) = \frac{1}{2x \cdot (1)} \cdot (2)$$

Simplifying, we have

$$ \frac{d}{dx} (\log_{e}{2x}) = \frac{1}{2x} \cdot (2)$$

$$ \frac{d}{dx} (\log_{e}{2x}) = \frac{2}{2x} $$

$$ \frac{d}{dx} (\log_{e}{2x}) = \frac{1}{x} $$

Step 5: Final Answer.

$$ \frac{d}{dx} (\ln{(2x)}) = \frac{1}{x}$$


Graph of ln(2x) vs. its derivative

Given the function

$$ f(x) = \ln{(2x)}$$

the graph is illustrated as

graph-of-fx-ln2x

And as we know by now, by deriving \(f(x) = \ln{(2x)}\), we get

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

which is illustrated graphically as

graph-of-derivarive of ln of 2x

Illustrating both graphs in one, we have

graph-of ln2x and its derivative

Analyzing the differences of these functions through these graphs, you can observe that the original function \(f(x) = \ln{(2x)}\) has a domain of

\( (0,\infty) \) or \( x | x > 0 \)

and exists within the range of

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

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

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

and exists within the range of

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

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


Examples

Below are some examples of deriving a natural logarthmic of a variable multiplied by two through the use of chain rule or general logarithmic derivative.

EXAMPLE 1

Derive: \(F(\beta) = \ln{(2\beta)}\)

Solution: Using chain rule, we have

Step 1: Analyze if the natural logarithm of \(2\beta\) 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

$$ u = g(\beta) = 2\beta $$

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

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

$$ g'(\beta) = \frac{d}{d\beta} 2\beta = 2 $$

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

$$ F'(\beta) = \left( \frac{1}{u} \right) \cdot (2) $$

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

$$ F'(\beta) = \left( \frac{1}{(2\beta)} \right) \cdot (2) $$

Step 6: Simplify and declare the final answer.

$$ F'(\beta) = \frac{2}{(2\beta)} $$

The final answer is:

$$ \frac{d}{d\beta} \ln{(2\beta)} = \frac{1}{(\beta)} $$

EXAMPLE 2

Derive: \(f(\lambda) = \ln{(2\lambda)}\)

Solution: Using the general logarithmic derivative formula, we have

Step 1: Express the function as \(f(\lambda) = \log_{e}{2\lambda}\) instead of \(\ln{(2\lambda)}\)

Step 2: Set the logarithmic argument as u.

$$ u = 2\lambda $$

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 \(f(\lambda) = \log_{e}{2\lambda}\) by applying the formula in step 2.

$$ \frac{d}{d\lambda} (\log_{e}{2\lambda}) = \frac{1}{u\ln{(e)}} \cdot \frac{d}{d\lambda} (u)$$

$$ \frac{d}{d\lambda} (\log_{e}{2\lambda}) = \frac{1}{2\lambda\ln{(e)}} \cdot \frac{d}{dx} (2\lambda)$$

$$ \frac{d}{d\lambda} (\log_{e}{2\lambda}) = \frac{1}{2\lambda \cdot (1)} \cdot (2)$$

$$ \frac{d}{d\lambda} (\log_{e}{2\lambda}) = \frac{2}{2\lambda}$$

$$ \frac{d}{d\lambda} (\log_{e}{2\lambda}) = \frac{1}{\lambda}$$

$$ \frac{d}{d\lambda} (\ln{(2\lambda)}) = \frac{1}{\lambda}$$


See also

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

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