The derivative of lnx is equal to 1/x. Ln(x) denotes the natural logarithm of x, that is, lnx= log_{e}x. Here we will find the derivative of ln(x) using the limit definition and chain rule of differentiation.

Note that lnx= log_{e}x. The derivative of lnx is denoted by d/dx(lnx), and its formula is given as follows:

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

## Find the Derivative of lnx

Answer: The derivative of lnx is 1/x. |

**Explanation:**

Let z= lnx

⇒ z = log_{e}x

⇒ e^{z} = x

Differentiating both sides we get

$e^z \dfrac{dz}{dx}=1$, obtained by the chain rule.

⇒ $e^{\ln x} \dfrac{dz}{dx}=1$

⇒ $x \dfrac{dz}{dx}=1$ as we know e^{lnx} =x.

⇒ $\dfrac{dz}{dx}=\dfrac{1}{x}$

That is, $\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$.

Therefore, the derivative of lnx is equal to 1/x, and this is obtained by the chain rule of differentiation.

**Also Read:**

## Derivative of lnx by First Principle

Let us recall the first principle of derivatives.

The derivative of a function f(x) by first principle is given by the following limit formula: $f'(x)$ = lim _{h→0} $\dfrac{f(x+h)-f(x)}{h}$. |

Put f(x) = lnx.

So (lnx)$’$ = lim_{h→0} $\dfrac{\ln (x+h)-\ln (x)}{h}$

= lim_{h→0} $\dfrac{\ln \Big(\dfrac{x+h}{x} \Big)}{h}$, by the difference rule of logarithms.

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

Let h/x = z, so that z→0 when h→0. Also, h=zx. |

= lim_{z→0} $\dfrac{\ln \big(1+z \big)}{zx}$

= $\dfrac{1}{x}$ lim_{z→0} $\dfrac{\ln \big(1+z \big)}{z}$

= $\dfrac{1}{x} \times 1$, using the limit formula: lim_{x→0} ln(1+x)/x =1.

= 1/x.

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

**More Reading:**

## FAQs

### Q1: What is the derivative of ln x?

Answer: The derivative of ln x is equal to 1/x.

### Q2: If y=lnx, then find dy/dx.

Answer: If y=lnx, then dy/dx =1/x.