The derivative of square root of logx is equal to 1/(2x root(logx)). In this post, we will learn how to differentiate root logx by the chain rule of derivatives. The formula for the derivative of root logx is given below.

$\dfrac{d}{dx}(\sqrt{\log x})$ $=\dfrac{1}{2x\sqrt{\log x}}$

## Derivative of Root logx by Chain Rule

By chain rule of derivatives, we will now prove that

$\dfrac{d}{dx}(\sqrt{\log x})$ $=\dfrac{1}{2x\sqrt{\log x}}$

*Proof:*

Let z=log x

Differentiating both sides with respect to x, we get that

$\dfrac{dz}{dx}=\dfrac{1}{x}$

Now, by the chain rule,

$\dfrac{d}{dx}(\sqrt{\log x})$ = $\dfrac{d}{dz}(\sqrt{z}) \times \dfrac{dz}{dx}$

= $\dfrac{d}{dz}(z^{1/2}) \times \dfrac{dz}{dx}$

= $\dfrac{1}{2} z^{\frac{1}{2}-1} \times \dfrac{1}{x}$ by the power rule of derivatives: d/dx(x^{n}) = nx^{n-1}

= $\dfrac{1}{2x} z^{-\frac{1}{2}}$

= $\dfrac{1}{2x z^{\frac{1}{2}}}$ using the rule of indices a^{-m}=1/a^{m}.

= $\dfrac{1}{2x \sqrt{z}}$

= $\dfrac{1}{2x \sqrt{\log x}}$ as z=logx.

So the derivative of root logx is equal to 1/(2x root(logx)) and this is derived by the chain rule of derivatives.

**Question:** Find the derivative of root logx at x=2.

*Answer:*

From the above derivative of root logx, we obtain that

$\Big[\dfrac{d}{dx}(\sqrt{\log x})\Big]_{x=2}$ $=\Big[\dfrac{1}{2x\sqrt{\log x}}\Big]_{x=2}$

= $\dfrac{1}{2 \times 2\sqrt{\log 2}}$

= $\dfrac{1}{4\sqrt{\log 2}}$

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## FAQs

**Q1: What is the derivative of root logx?**

**Answer:** The derivative of root log x is 1/(2x root(logx)).