Derivative of log 4x | Derivative of ln 4x

The logarithm of 4x is denoted by log(4x) and its derivative is 1/x if the base is e. In this post, we will learn how to find the derivative of log(4x) with any base. The following methods will be used:

• Implicit function differentiation
• First principle of derivatives
• Chain rule of derivatives.

Derivative of log 4x Formula

The derivative of log 4x with base a is equal to 1/(x ln a). So the derivative of log 4x is 1/(x log_e10) if the default base is 10.

The formulae for the derivatives of log 4x with different bases are given in the table below:

Log Functions	Derivative
loga 4x	1/(x logea)
log10 4x	1/(x loge10)
loge 4x	1/x

Derivative of ln 4x

The natural logarithm of 4x is the logarithm of 4x with base e, and it is denoted by ln(4x). Let us now find its derivative.

Question: What is the Derivative of ln 4x?

Answer: The derivative of ln 4x is 1/x.

Proof:

Note that ln 4x = loge 4x

∴ d/dx(ln 4x) = d/dx(loge 4x)

As we know that d/dx(loga 4x)= 1/(x loge a), we get

d/dx(loge 4x) = 1/(x loge e) = 1/x as ln e =1.

∴ The derivative of ln 4x is 1/x.

What is the Derivative of log 4x?

Answer: The derivative of log_a(4x) is 1/(x log_ea).

Proof: We will use the implicit function differentiation method.

Let y = log_a4x.

By the properties of logarithms, we have

a^y = 4x

Differentiating both sides with respect to x, we get that

a^y log_ea $\frac{dy}{dx}$ = 4

⇒ 4x log_ea $\frac{dy}{dx}$ = 4 as we know $a^{\log_a {4x}}=4x$

⇒ $\frac{dy}{dx}$ = 1/(x log_ea).

This shows that the derivative of log_a 4x is 1/(x log_ea), obtained by the implicit differentiation method.

Also Read:

Derivative of esin x: The derivative of esin x is cos x esin x. Integration of modulus of x: The integration of mod x is -x|x|/2+c. Derivative of 1/x: The derivative of 1 by x is -1/x2.

Derivative of log 4x from First Principle

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

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

Put f(x) = log_a4x.

So the derivative of log_a4x using the first principle is

$\dfrac{d}{dx}(\log_a 4x)$ $=\lim\limits_{h \to 0}\dfrac{\log_a 4(x+h)- \log_a 4x}{h}$

$=\lim\limits_{h \to 0}\dfrac{\log_a \frac{4x+4h}{4x}}{h}$ by the logarithm rule $\log_a x -\log_a y$ $=\log_a \frac{x}{y}$

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

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

[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.

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

Thus, the derivative of log of 4x with base a is 1/(x log_ea) and this is obtained from the first principle of derivatives.

Derivative of log 4x by Chain Rule

Let f=log_a 4x and u = 4x.

∴ df/dx = d/dx(log_a 4x)

By the chain rule of derivatives, we have

df/dx = d/du(log_a u) × du/dx

= 1/(u log_ea) × 4

= 1/(4x log_ea) × 4 as u=4x

= 1/(x log_ea)

So the derivative of log 4x with base a is 1/(x log_ea) which is achieved by the chain rule of differentiation. Putting a=e, we get the derivative of natural log of 4x which is d/dx(log_e 4x) = 1/x as we know that log_ee = 1.

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