Derivative of root tanx by First Principle

The derivative of root tanx is equal to sec²x/(2√tanx). In this post, we will learn how to differentiate the square root of tan x using the first principles.

Derivative of square root of tanx from first principle

Question: Find the derivative of $\sqrt{\tan x}$ from first principle.

Solution:

The below steps have to be followed to find the derivative of square root of tanx using first principle.

Step 1:

Let f(x) = $\sqrt{\tan x}$

We know that the derivative of f(x) by first principle is given by

$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$.

Step 2:

Putting f(x)=$\sqrt{\tan x}$ in the above limit, we get that

$\dfrac{d}{dx}(\sqrt{\tan x})$ $=\lim\limits_{h \to 0} \dfrac{\sqrt{\tan (x+h)}-\sqrt{\tan x}}{h}$

We will multiply both the numerator and the denominator by √tan (x + h) + √tan x. By doing so, the above

= $\lim\limits_{h \to 0} \Big[ \dfrac{\sqrt{\tan (x+h)}-\sqrt{\tan x}}{h}$ $\times \dfrac{\sqrt{\tan (x+h)}+\sqrt{\tan x}}{\sqrt{\tan (x+h)}+\sqrt{\tan x}} \Big]$

= $\lim\limits_{h \to 0} \dfrac{(\sqrt{\tan (x+h)})^2-(\sqrt{\tan x})^2}{h(\sqrt{\tan (x+h)}+\sqrt{\tan x})}$, this is obtained by using the formula (a-b)(a+b)=a²-b².

= $\lim\limits_{h \to 0} \Big[ \dfrac{\tan (x+h)-\tan x}{h}$ $\times \dfrac{1}{\sqrt{\tan (x+h)}+\sqrt{\tan x}} \Big]$ …(∗)

Step 3:

Now, by the first principle of derivatives, the limit $\lim\limits_{h \to 0} \dfrac{\tan (x+h)-\tan x}{h}$ $=\dfrac{d}{dx}(\tan x)$ = sec²x.

So applying the product rule of limits, we deduce from (∗) that

$\dfrac{d}{dx}(\sqrt{\tan x})$ = $\lim\limits_{h \to 0} \dfrac{\tan (x+h)-\tan x}{h}$ $\times \lim\limits_{h \to 0} \dfrac{1}{\sqrt{\tan (x+h)}+\sqrt{\tan x}}$

= sec²x × $\dfrac{1}{\sqrt{\tan (x+0)}+\sqrt{\tan x}}$

= sec²x × $\dfrac{1}{2\sqrt{\tan x}}$

= $\dfrac{\sec^2 x}{2\sqrt{\tan x}}$

Thus, the derivative of square root of tanx is equal to sec²x/(2√tanx) and this is obtained from the first principle of derivatives.

Related Topics:

Derivative of $\sqrt{\sin x}$ from first principle

Derivative of $\sqrt{\cos x}$ from first principle

Derivative of $\sqrt{\sin x}$ by chain rule

Derivative of $\sqrt{\cos x}$ by chain rule

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