Derivative of log 3x | Derivative of ln 3x

The function “log” usually denotes the common logarithm, that is, it is a logarithm with base 10. If no base is mentioned, then one should consider the default base 10. So we have log x = log₁₀x. Note that the logarithm with base e is called the natural logarithm, denoted by ln. That is, log_ex = ln x. In this post, we will find the derivative of log x with respect to x using the following methods:

• By chain rule of derivatives.
• By first principle of derivatives.
• By the method of derivatives of implicit functions.

What is the derivative of log 3x?

Answer: The derivative of log_a(3x) is 1/(x log_ea).

That is, the derivative of log 3x with base a is equal to 1/(x ln a). So the derivative of log 3x is 1/(x log_e10) if the default base is 10.

The formulae for the derivatives of log 3x with different bases are given in the table below:

Log Functions	Derivative
loga 3x	1/(x logea)
log10 3x	1/(x loge10)
loge 3x	1/x

Also Read:

Derivative of esin x : The derivative of esin x is cos x esin x. Derivative of 1/x : The derivative of 1 by x is -1/x2.

Derivative of log 3x by Chain Rule

The chain rule of differentiation says that if f is a function of u and u is a function of x, then the derivative of f with respect to x is equal to

$\dfrac{df}{dx}=\dfrac{df}{du} \cdot \dfrac{du}{dx}.$

Now, we will find the derivative of log_a3x (log of 3x with base a) by the chain rule of derivatives.

Note that f=log_a 3x is a function of u=3x. So we take

$u = 3x \quad \Rightarrow \dfrac{du}{dx} = 3.$

By the above chain rule of derivatives, the derivative of log 3x is

$\dfrac{d}{dx}(\log_a 3x)=\dfrac{d}{dx}(\log_a u)$ $=\dfrac{d}{du}(\log_a u) \cdot \dfrac{du}{dx}$

$=\dfrac{1}{u \log_e a} \cdot 3$ as the derivative of log_ax is 1/(x log_ea).

$=\dfrac{1}{3x \log_e a}\cdot 3$ as u=3x.

$=\dfrac{1}{x \log_e a}$.

So the derivative of log 3x with base a is 1/(x log_ea). We achieve this by the chain rule of differentiation. Putting a=e, we get the derivative of natural log of 3x which is d/dx(log_e 3x) = 1/x as we know that log_ee = 1.

Derivative of ln 3x

Question: What is the Derivative of ln 3x?

Answer: The derivative of ln 3x is 1/x.

Proof:

Note that ln 3x = loge 3x

∴ d/dx(ln 3x) = d/dx(loge 3x)

As we know that d/dx(loga 3x)= 1/(x loge a), we get

d/dx(loge 3x) = 1/(x loge e) = 1/x as ln e =1.

Derivative of log 3x from First Principle

The derivative of a function f(x) by first principle is given by the following limit:

$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$

Take $f(x)=\log_a 3x$ in the above formula. So the derivative of log of 3x with base a using the first principle is

$\dfrac{d}{dx}(\log_a 3x)$ $=\lim\limits_{h \to 0}\dfrac{\log_a 3(x+h)- \log_a 3x}{h}$

$=\lim\limits_{h \to 0}\dfrac{\log_a \frac{3x+3h}{3x}}{h}$ using the formula of $\log_a x -\log_a y$ $=\log_a \frac{x}{y}$

$=\lim\limits_{h \to 0}\dfrac{\log_a \left(1+\frac{h}{x} \right)}{h}$

$=\lim\limits_{h \to 0}\dfrac{\log_a \left(1+\frac{h}{x} \right)}{h/x}$ $\times \dfrac{1}{x}$

$=\dfrac{1}{x}\lim\limits_{h \to 0}\dfrac{\log_a \left(1+\frac{h}{x} \right)}{h/x}$

[Let t=h/x. Then t→0 as h→0]

$=\dfrac{1}{x}\lim\limits_{t \to 0}\dfrac{\log_a \left(1+t \right)}{t}$

$=\dfrac{1}{x} \times \log_e a$ as the limit of log_a(1+t) / t is log_ea when t→0.

$=\dfrac{1}{x\log_e a}$

Hence the derivative of log_a 3x is 1/(x log_ea) and this is obtained from the first principle of derivatives.

Derivative of log 3x by Implicit Differentiation

Prove that d/dx(log_a3x) = 1/(x log_ea) by the method of differentiation for implicit functions.

Proof:

Let y = log_a3x. By the properties of logarithms, we have

a^y = 3x

Differentiating with respect to x, we get that

a^y log_ea $\frac{dy}{dx}$ = 3

⇒ 3x log_ea $\frac{dy}{dx}$ = 3 as we know $a^{\log_a {3x}}=3x$

⇒ $\frac{dy}{dx}$ = 1/(x log_ea).

Thus we have shown that the derivative of log_a 3x is 1/(x log_ea) by implicit differentiation method.

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