The derivative of tan root x is equal to (sec^{2}√x)/2√x. By the chain rule of differentiation, the derivative formula of tan(√x) with respect to x is given as follows:

$\dfrac{d}{dx}(\tan \sqrt{x})=\dfrac{\sec^2 \sqrt{x}}{2\sqrt{x}}$.

## Derivative of tan(√x) by Chain Rule

**Answer:** The derivative of tan√x is (sec^{2}√x)/2√x.

**Explanation:**

Let us put

z = √x.

Differentiating both sides w.r.t. x, we get that

dz/dx = d/dx(x^{1/2}) = 1/2 x^{1-1/2} = 1/(2√x) [by the power rule of differentiation].

Now, the given derivative

$\dfrac{d}{dx}(\tan \sqrt{x})$

= $\dfrac{d}{dx}(\tan z)$

= $\dfrac{d}{dz}(\tan z) \times \dfrac{dz}{dx}$, by the chain rule of differentiation.

= $\sec^2 z \times \dfrac{1}{2\sqrt{x}}$ as we have dz/dx = 1/(2√x).

= $\dfrac{\sec^2 \sqrt{x}}{2\sqrt{x}}$, putting back the value of z=√x.

So the derivative of tan root x is equal to (sec^{2}√x)/2√x, and this is proved by the chain rule of differentiation.

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## FAQs

### Q1: What is the derivative of tan root x?

Answer: The derivative of tan root x is (sec^{2}√x)/2√x, that is, d/dx (tan√x) = (sec^{2}√x)/2√x.

### Q2: If y= tan root x, then find dy/dx.

Answer: If y=tan√x, then dy/dx = (sec^{2}√x)/2√x.