Derivative of Root cotx: Proof by First Principle, Chain Rule

The derivative of root tanx is equal to -cosec²x/(2√cotx). Here we will differentiate square root of cot x by chain rule and also using the first principle of derivatives.

The derivative formula of root cotx is given by

$\dfrac{d}{dx}(\sqrt{\cot x})=-\dfrac{\text{cosec}^2 x}{2\sqrt{\cot x}}$.

Derivative of square root of cotx from first principle

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

Solution:

We will follow the below steps to find the derivative of square root of cot x using first principle. The steps are as follows.

Step 1:

Recall, the derivative of a function f(x) by first principle. It is given by the limit below:

$\dfrac{d}{dx}(f(x))$ = lim_h→0 $\dfrac{f(x+h)-f(x)}{h}$.

Step 2:

Put f(x)=$\sqrt{\cot x}$.

So the derivative of square root cotx will be equal to

$\dfrac{d}{dx}(\sqrt{\cot x})$ = lim_h→0 $\dfrac{\sqrt{\cot (x+h)}-\sqrt{\cot x}}{h}$

Multiply both the numerator and the denominator by √cot(x + h) + √cotx. This will give us

$\dfrac{d}{dx}(\sqrt{\cot x})$ = lim_h→0 $\Big[ \dfrac{\sqrt{\cot (x+h)}-\sqrt{\cot x}}{h}$ $\times \dfrac{\sqrt{\cot (x+h)}+\sqrt{\cot x}}{\sqrt{\cot (x+h)}+\sqrt{\cot x}} \Big]$

= lim_h→0 $\dfrac{(\sqrt{\cot (x+h)})^2-(\sqrt{\cot x})^2}{h(\sqrt{\cot (x+h)}+\sqrt{\cot x})}$, using the algebraic identity (a-b)(a+b)=a²-b².

= lim_h→0 $\Big[ \dfrac{\cot (x+h)-\cot x}{h}$ × $\dfrac{1}{\sqrt{\cot (x+h)}+\sqrt{\cot x}} \Big]$ …(∗)

Step 3:

Note that the limit $\lim\limits_{h \to 0} \dfrac{\cot (x+h)-\cot x}{h}$ is equal to $\dfrac{d}{dx}(\cot x)$, by the first principle of derivatives, which is equal to -cosec²x.

Now, from (∗) we deduce (applying product rule) that

$\dfrac{d}{dx}(\sqrt{\cot x})$ = lim_h→0 $\dfrac{\cot (x+h)-\cot x}{h}$ × lim_h→0 $\dfrac{1}{\sqrt{\cot (x+h)}+\sqrt{\cot x}}$

= -cosec²x $\times \dfrac{1}{\sqrt{\cot (x+0)}+\sqrt{\cot x}}$

= -cosec²x $ \times \dfrac{1}{2\sqrt{\cot x}}$

= $-\dfrac{\text{cosec}^2 x}{2\sqrt{\cot x}}$

Thus, the derivative of square root of cotx by the first principle is equal to -cosec²x/(2√cotx).

Related Topics:

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

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

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

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

Derivative of root cotx by chain rule

Let u=cotx.

So du/dx = -cosec²x.

Now, by the chain rule, the derivative of square root of cotx is equal to

$\dfrac{d}{dx}(\sqrt{\cot x})$ = $\dfrac{d}{dx}(\sqrt{u})$

= $\dfrac{d}{du}(\sqrt{u}) \times \dfrac{du}{dx}$

= $\dfrac{d}{du}$(u^1/2) × -cosec²x

= 1/2 u^1/2-1 × -cosec²x

= 1/(2√u) × -cosec²x

= – cosec²x/(2√cotx).

So the derivative of square root of cotx by the chain rule is equal to -cosec²x/(2√cotx).

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

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