Derivative of log 2x | Derivative of ln 2x

The “log” function denotes a logarithm with base 10 referred to as common logarithms. So log 2x = log₁₀ 2x. If the base is e, then log_e 2x = ln 2x is called the natural logarithm. Here, we will find the derivative of log 2x and ln 2x with respect to x. We have:

• The derivative of log 2x is 1/(x log_e10).
• The derivative of log_a 2x is 1/(x log_ea). Here the base is a.
• The derivative of ln 2x is 1/x.

What is the derivative of log 2x?

We know that the derivative of log_a(2x) is 1/(x log_ea), that is,

d/dx{log_a(2x)} = 1/(x log_ea) = 1/(x ln a).

So the derivative of log 2x is 1/(x log_e10) where the base is 10.

The formulae for the derivatives of log 2x with different bases are given in the table below:

Log Functions	Derivative
loga 2x	1/(x logea)
log10 2x	1/(x loge10)
loge 2x	1/x

Derivative of ln 2x

Question: What is the Derivative of ln 2x?

Answer: The derivative of ln 2x is 1/x.

Proof:

Note that ln 2x = loge 2x

∴ d/dx(ln 2x) = d/dx(loge 2x)

As we know that d/dx(loga 2x)= 1/(x loge a), we get

d/dx(loge 2x) = 1/(x loge e) = 1/x as ln e =1.

Also Read:

Derivative of esin x : The derivative of esin x is cos x esin x. Derivative of ln 3x : The derivative of ln 3x is 1/x. Derivative of sin 3x : The derivative of sin 3x is 3cos 3x. Derivative of 1/x : The derivative of 1 by x is -1/x2.

Derivative of log 2x by Chain Rule

If f is a function of u and u is a function of x, then the derivative of f with respect to x by the chain rule is equal to

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

Take f(x) = log₁₀ 2x (log of 2x with base 10).

Here u=2x ⇒ du/dx =2.

The derivative of log 2x by the chain rule is

$\dfrac{d}{dx}(\log_{10} 2x)=\dfrac{d}{dx}(\log_{10} u)$ $=\dfrac{d}{du}(\log_{10} u) \cdot \dfrac{du}{dx}$

= $\dfrac{1}{u \log_{e} 10} \cdot 2$ as the derivative of log₁₀ x is 1/(x log_e10).

= $\dfrac{1}{2x \log_e 10}\cdot 2$ as u=2x.

= $\dfrac{1}{x \log_e 10}$.

So the derivative of log 2x is 1/(x log_e10). Now, to obtain the derivative of ln 2x, put a=e. Thus, the derivative of the natural log of 2x is d/dx(log_e 2x) = 1/x as we know that log_ee = 1.

Derivative of log 2x from First Principle

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

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

Take f(x)=log₁₀ 2x in the above formula. So the derivative of log of 2x with base 10 by the first principle is

$\dfrac{d}{dx}(\log_{10} 2x)$ $=\lim\limits_{h \to 0}\dfrac{\log_{10} 2(x+h)- \log_{10} 2x}{h}$

$=\lim\limits_{h \to 0}\dfrac{\log_{10} \frac{2x+2h}{2x}}{h}$. Here, we have used the formula: $\log_a x -\log_a y$ $=\log_a \frac{x}{y}$

$=\lim\limits_{h \to 0}\dfrac{\log_{10} \left(1+\frac{h}{x} \right)}{h}$

$=\lim\limits_{h \to 0}\dfrac{\log_{10} \left(1+\frac{h}{x} \right)}{h/x}$ $\times \dfrac{1}{x}$

$=\dfrac{1}{x}\lim\limits_{h \to 0}\dfrac{\log_{10} \left(1+\frac{h}{x} \right)}{h/x}$

[Let t=h/x. Then t→0 as h→0]

$=\dfrac{1}{x}\lim\limits_{t \to 0}\dfrac{\log_{10} \left(1+t \right)}{t}$

$=\dfrac{1}{x} \times \log_e 10$ as the limit of log_a(1+t) / t is log_ea when t→0.

$=\dfrac{1}{x\log_e 10}$

Hence the derivative of log₁₀ 2x is 1/(x log_e10) and this is obtained from the first principle of derivatives.

Derivative of log 2x by Implicit Differentiation

Prove that d/dx(log₁₀2x) = 1/(x log_e10) by the method of differentiation for implicit functions.

Proof:

Let y = log₁₀2x. Using the properties of logarithms, we have

10^y = 2x

Differentiating with respect to x, we get that

10^y log_e10 $\frac{dy}{dx}$ = 2

⇒ 2x log_e10 $\frac{dy}{dx}$ = 2 as we know $a^{\log_a {2x}}=2x$

⇒ $\frac{dy}{dx}$ = 1/(x log_e10).

Thus we have shown that the derivative of log₁₀ 2x is 1/(x log_e10) by the implicit differentiation method.

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