Derivative of ln(3x) with Proofs and Graphs

The natural logarithm of 3x, also denoted as ln(3x), is the logarithm of 3x to base e (euler’s number)The derivative of the natural logarithm of 3x is equal to one over x, 1/x. 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(3x). We will review some principles, definitions, formulas, graph comparisons of underived and derived ln(3x), proofs, derivation techniques, and examples.

Derivative of natural log of 3x

Relevant for

Learning how to find the derivative of the natural logarithm of 3x.

See proofs

Derivative of natural log of 3x

Relevant for

Learning how to find the derivative of the natural logarithm of 3x.

See proofs

Proofs of the Derivative of Natural Logarithmic of 3x

You may be wondering why \( \ln{(x)} \), \( \ln{(x)} \), \( \ln{(3x)} \), or even \( \ln{(nx)} \) have the same derivatives. To show why, listed below are some proofs of the derivative of \( \ln{(3x)} \).

Aside from being proofs, these can also serve as the main methods of deriving 3x.

Proof of the derivative of ln(3x) using the Chain Rule Formula

In the derivative of the natural log of 3x, can be found using the chain rule. The natural logarithmic function will be the outer function f(u) in the composite function ln(3x), whereas the monomial 3x 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: Derivative of Natural log (ln(x)).

Let’s have the derivative of the function

$$ F(x) = \ln{(3x)}$$

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.

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


$$ u = 3x$$

Setting the monomial 3x as the inner function of f(u) by denoting it as g(x), we have

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

$$ g(x) = 3x$$

$$ u = g(x)$$

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

$$ g(x) = 3x$$

$$ g'(x) = 3$$

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

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

$$ \frac{dy}{dx} = \left(\frac{1}{u} \right) \cdot (3)$$

Substituting u into f'(u), we have

$$ \frac{dy}{dx} = \left(\frac{1}{(3x)} \right) \cdot (3)$$

$$ \frac{dy}{dx} = \frac{3}{3x}$$

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

As a result, we arrive at the ln(3x) derivative formula.

$$ \frac{d}{dx} \ln{(3x)} = \frac{1}{x}$$

Proof of the derivative of ln(3x) using implicit differentiation

You are advised to learn/review the derivatives of exponential functions and implicit differentiation for this proof.

Given that the equation

$$ y = \ln{(3x)}$$

In general logarithmic form, it is

$$ \log_{e}{3x} = y$$

And in exponential form, it is

$$ e^y = 3x$$

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

$$ e^y = 3x$$

$$ \frac{d}{dx} (e^y) = \frac{d}{dx} (3x) $$

$$ e^y \cdot \frac{dy}{dx} = 3 $$

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

$$ \frac{dy}{dx} = \frac{3}{e^y} $$

We recall that in the beginning, \( y = \ln{(3x)} \). Substituting this to the y of our derivative, we have

$$ \frac{dy}{dx} = \frac{3}{e^{(\ln{(3x)})}} $$

Simplifying and applying a property of logarithm, we have

$$ \frac{dy}{dx} = \frac{3}{3x} $$

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

Evaluating, we now have the derivative of \( y = \ln{(3x)} \)

$$ y’ = \frac{1}{x} $$

Other Methods to derive the Natural Logarithmic of 3x

Aside from the two proofs mentioned above, which also serve as the primary methods for deriving ln(3x), another method can be utilized to derive this function.

Derivative of the natural logarithmic of 3x by using the derivative of general logarithmic function.

Step 1: Express the function as \(f(x) = \log_{e}{3x}\) instead of \(\ln{(3x)}\)

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 \(f(x) = \log_{e}{3x}\) 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 3x. Hence,

$$ \frac{d}{dx} (\log_{e}{3x}) = \frac{1}{3x\ln{(e)}} \cdot \frac{d}{dx} (3x)$$

$$ \frac{d}{dx} (\log_{e}{3x}) = \frac{1}{3x\ln{(e)}} \cdot (3)$$

Step 4: Algebraically evaluate and simplify.

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

$$ \frac{d}{dx} (\log_{e}{3x}) = \frac{1}{3x \cdot (1)} \cdot (3)$$

Simplifying, we have

$$ \frac{d}{dx} (\log_{e}{3x}) = \frac{1}{3x} \cdot (3)$$

$$ \frac{d}{dx} (\log_{e}{3x}) = \frac{3}{3x} $$

$$ \frac{d}{dx} (\log_{e}{3x}) = \frac{1}{x} $$

Step 5: Final Answer.

$$ \frac{d}{dx} (\ln{(3x)}) = \frac{1}{x}$$

Graph of ln(3x) vs. its derivative

In the instance of this function

$$ f(x) = \ln{(3x)}$$

the graph is illustrated as


And as we learned above, deriving \(f(x) = \ln{(3x)}\) will be

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

which is illustrated graphically as

graph-of-derivative of-ln3x

Illustrating both graphs in one, we have

graph-of-ln3x and its derivative

By examining the differences between these functions using these graphs, you can see that the original function \(f(x) = \ln{(3x)}\) has a domain of

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

and lies 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 \( x | x \neq 0 \)

which lies within the range of

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


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


Derive: \(F(\beta) = \ln{(3\beta)}\)

Solution: Using chain rule, we have

Step 1: Analyze if the natural logarithm of \(3\beta\) 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

$$ u = g(\beta) = 3\beta $$

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

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

$$ g'(\beta) = \frac{d}{d\beta} 3\beta = 3 $$

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

$$ F'(\beta) = \left( \frac{1}{u} \right) \cdot (3) $$

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

$$ F'(\beta) = \left( \frac{1}{(3\beta)} \right) \cdot (3) $$

Step 6: Simplify and declare the final answer.

$$ F'(\beta) = \frac{3}{(3\beta)} $$

The final answer is:

$$ \frac{d}{d\beta} \ln{(3\beta)} = \frac{1}{(\beta)} $$


Derive: \(f(\lambda) = \ln{(3\lambda)}\)

Solution: Using the general logarithmic derivative formula, we have

Step 1: Express the function as \(f(\lambda) = \log_{e}{3\lambda}\) instead of \(\ln{(3\lambda)}\)

Step 2: Set the logarithmic argument as u.

$$ u = 3\lambda $$

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 \(f(\lambda) = \log_{e}{3\lambda}\) by applying the formula in step 2.

$$ \frac{d}{d\lambda} (\log_{e}{3\lambda}) = \frac{1}{u\ln{(e)}} \cdot \frac{d}{d\lambda} (u)$$

$$ \frac{d}{d\lambda} (\log_{e}{3\lambda}) = \frac{1}{3\lambda\ln{(e)}} \cdot \frac{d}{dx} (3\lambda)$$

$$ \frac{d}{d\lambda} (\log_{e}{3\lambda}) = \frac{1}{3\lambda \cdot (1)} \cdot (3)$$

$$ \frac{d}{d\lambda} (\log_{e}{3\lambda}) = \frac{3}{3\lambda}$$

$$ \frac{d}{d\lambda} (\log_{e}{3\lambda}) = \frac{1}{\lambda}$$

$$ \frac{d}{d\lambda} (\ln{(3\lambda)}) = \frac{1}{\lambda}$$

See also

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