Derivative of ln4x | ln(4x) Derivative

The derivative of ln4x is equal to 1/x. The derivative of ln4x is denoted by d/dx(ln4x), and its formula is given by

$\dfrac{d}{dx}(\ln 4x)=\dfrac{1}{x}$.

Find the Derivative of ln4x

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

Explanation:

As ln4x = ln4 + lnx, the derivative of ln 4x is equal to

$\dfrac{d}{dx}$ (ln 4x) = $\dfrac{d}{dx}$(ln4 + lnx)

= $\dfrac{d}{dx}$ (ln 4) + $\dfrac{d}{dx}$ (ln x), by the sum rule of derivatives

= 0 + $\dfrac{1}{x}$ as ln4 is a constant and the derivative of a constant is 0.

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

So the derivative of ln4x is equal to 1/x.

Also Read: Derivative of lnx by First Principle

Derivative of ln(x+1) | Derivative of ln2x

Derivative of ln4x by First Principle

The derivative of f(x) by first principle is given by the formula $\dfrac{d}{dx}(f(x))$ = limh→0 $\dfrac{f(x+h)-f(x)}{h}$.

Let f(x) = ln4x.

So the derivative of ln4x by the first principle is

= lim_h→0 $\dfrac{\ln 4(x+h)-\ln 4x}{h}$

= lim_h→0 $\dfrac{\ln \Big(\dfrac{4x+4h}{4x} \Big)}{h}$, using the quotient rule of logarithms.

= lim_h→0 $\dfrac{\ln \Big(1+\dfrac{h}{x} \Big)}{h}$

Let h/x = t, so that t→0 when h→0. Also, h=tx.

= lim_t→0 $\dfrac{\ln \big(1+t \big)}{tx}$

= $\dfrac{1}{x}$ lim_t→0 $\dfrac{\ln \big(1+t \big)}{t}$

= $\dfrac{1}{x}$ × 1, as lim_x→0 ln(1+x)/x = 1.

= 1/x.

So the derivative of ln4x by the first principle is equal to is 1/x.

Read Also: Derivative of 1/lnx

What is the derivative of lnx²

Derivative of ln4x by Chain Rule

Let z=4x.

∴ dz/dx = 4.

So the derivative of ln4x by the chain rule is

$\dfrac{d}{dx}(\ln 4x)$

= $\dfrac{d}{dz}(\ln z) \times \dfrac{dz}{dx}$

= $\dfrac{1}{z}$ × 4

= $\dfrac{4}{4x}$ as z=4x.

= 1/x.

So the derivative of ln4x by the chain rule is 1/x.

Find the derivative of sin²x

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