The derivative of x^1/x (x to the power 1/x) is denoted by d/dx(x^{1/x}) and its value is equal to x^{1/x}[1/x^{2} – 1/x^{2} ⋅ log_{e}x].

The derivative formula of x^{1/x} is given below:

d/dx(x^{1/x}) = x^{1/x -2} [1 – log_{e}x].

Let us now learn how to differentiate x^{1/x}.

## Derivative of x to the power 1/x

Question: Prove that d/dx(x^{1/x}) = x^{1/x -2} [1 – log_{e}x]. |

**Answer:**

Let us put

y=x^{1/x}.

Here we need to find dy/dx. Taking logarithms on both sides, we get that

log_{e} y = log_{e} x^{1/x}

⇒ log_{e}y = 1/x ⋅ log_{e}x, as we know the logarithm formula log_{a}b^{n} = n log_{a}b.

Differentiating both sides w.r.t x, we have

$\dfrac{d}{dx}(\log_e y)=\dfrac{d}{dx}(\dfrac{1}{x} \log_e x)$

⇒ $\dfrac{1}{y} \dfrac{dy}{dx}$ $=\dfrac{1}{x} \dfrac{d}{dx}(\log_e x)+\log_e x\dfrac{d}{dx}(\dfrac{1}{x})$, by the product rule of derivatives.

⇒ $\dfrac{1}{y} \dfrac{dy}{dx}$ $=\dfrac{1}{x} \cdot \dfrac{1}{x}+\log_e x \cdot \dfrac{-1}{x^2}$ as we know d/dx(log_{e}x) =1/x and the derivative of 1/x is equal to -1/x^{2}.

⇒ $\dfrac{dy}{dx}=y(\dfrac{1}{x^2}-\dfrac{1}{x^2\log_e x}$

= $x^{\frac{1}{x}} \dfrac{1}{x^2} \Big(1-\log_e x \Big)$ as y=x^{1/x}.

= x^{1/x -2} [1 – log_{e}x].

So the derivative of x^{1/x} (x to the power 1/x) is equal to x^{1/x -2} [1 – log_{e}x], and it is obtained by the logarithmic differentiation.

**More Derivatives:**

Derivative of x^{x} | Derivative of x^{sinx}

Find the derivative of x^{tanx}

Derivative of sinx/x | Derivative of x^{logx}

## FAQs

### Q1: What is the derivative of x^1/x?

**Answer:** The derivative of x^1/x (x raised to the power tanx) is equal to x^{1/x -2} [1 – log_{e}x].