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

**Relevant for**…

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

##### CALCULUS

**Relevant for**…

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

## 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

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

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

which is illustrated graphically as

Illustrating both graphs in one, we have

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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