Derivative of Square Root of x | Root x Derivative

Derivative of root x. The square root of x is an important function in mathematics. So it is natural to study the derivative of the square root of x. We will use the formula of power rule of derivatives to find it. We will also evaluate the derivative of the square root of x by the limit definition.

Power rule of derivatives: d/dx(xⁿ)=nx^n-1

What is the derivative of square root of x?

Step 1: We rewrite root x using the rule of indices.
\[\sqrt{x}=x^{1/2}\]

Step 2: Apply the above power rule of derivatives.

$\frac{d}{dx}(\sqrt{x})$ $=\frac{d}{dx}(x^{1/2})$ $=\frac{1}{2}x^{1/2-1}$

Step 3: Simplify the expression.

$\therefore \dfrac{d}{dx}(\sqrt{x})=\dfrac{1}{2}x^{-1/2}$

$=\dfrac{1}{2} \times \dfrac{1}{x^{1/2}}$

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

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

So the derivative of the square root of $x$ by the power ruleof derivatives is equal to $\frac{1}{2\sqrt{x}}$, that is,

\[\dfrac{d}{dx}(\sqrt{x})=\dfrac{1}{2\sqrt{x}}.\]

Derivative of esin x : The derivative of esin x is cos x esin x. Integration of mod x : The integration of mod x is -x|x|/2+c.

Alternative Method: Next, we will find the derivative of x^1/2 by the substitution method.

Let $y=\sqrt{x}.$

We take squares on both sides. Thus we get

$y^2=x$

Differentiating with respect to $x$,

$2y \dfrac{dy}{dx}=1$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2y}$

Putting the value of y, that is, y=√x we obtain that

$\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}.$

In other words, we can say that

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

Now, we will find the derivative of root x using first principle.

Derivative of Square Root of x from First Principle

Derivative of root x by first principle. Let f(x)=√x. We need to find the derivative of $f(x).$ From first principle of derivatives, we have

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

$=\lim\limits_{h \to 0} \dfrac{\sqrt{x+h}-\sqrt{x}}{h}$

$=\lim\limits_{h \to 0}[\frac{\sqrt{x+h}-\sqrt{x}}{h} \times \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}]$

$=\lim\limits_{h \to 0} \dfrac{x+h-x}{h \sqrt{x+h}+\sqrt{x}}$

$=\lim\limits_{h \to 0}\dfrac{h}{h\sqrt{x+h}+\sqrt{x}}$

$=\lim\limits_{h \to 0}\dfrac{1}{\sqrt{x+h}+\sqrt{x}}$

$=\dfrac{1}{\sqrt{x+0}+\sqrt{x}}$

$=\dfrac{1}{\sqrt{x}+\sqrt{x}}$

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

So the derivative of square root of x by the first principle is $\dfrac{1}{2\sqrt{x}}$.

Derivative of cube root of x: The derivative of the cube root of x is 1/(3x^{2/3}) Integration of root x: The integration of  √x is 2/3x^{3/2}

Derivative of Root x by Logarithmic Differentiation

Now, we will find the derivative of √x with the help of the logarithmic derivative. Write

y= √x

⇒ y = x^1/2

Taking natural logarithm (with base e) of both sides, we get that

ln y = 1/2 ln x

Differentiating with respect to x, we have

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{x}$

⇒ $\frac{dy}{dx} =\frac{y}{2x}$ $= \frac{\sqrt{x}}{2x}$ $=\frac{1}{2\sqrt{x}}$.

So by the logarithmic differentiation, the derivative of root x is $\frac{1}{2\sqrt{x}}$.

Application of Derivative of square root of x

We have evaluated the derivative of root $x$ above and get that

$\frac{d}{dx}(\sqrt{x})$ $=\frac{1}{2\sqrt{x}}$ $\cdots (i)$

Using this fact, we can find the derivatives of many functions involving square roots using the chain rule of derivatives. The derivative of the square root of 3x+2 will be calculated as follows:

Example 1: Find $\dfrac{d}{dx}(\sqrt{3x+2})$

Let $z=3x+2$

∴ $\dfrac{d}{dx}(\sqrt{3x+2})$ $=\dfrac{d}{dx}(\sqrt{z})$

$=\dfrac{d}{dz}(\sqrt{z}) \cdot \dfrac{dz}{dx}$ (by the chain rule)

$=\dfrac{1}{2\sqrt{z}} \cdot \dfrac{d}{dx}(3x+2)$ by Equation (i)

$=\dfrac{1}{2\sqrt{z}} \cdot 3$

$=\dfrac{3}{2\sqrt{3x+2}}$  $[\because z=3x+2]$

We calculate the derivative of the square root of sin x below.

Example 2: Find $\dfrac{d}{dx}(\sqrt{\sin x})$

Let $z=\sin x$

∴ $\dfrac{d}{dx}(\sqrt{\sin x})$ $=\dfrac{d}{dx}(\sqrt{z})$

$=\dfrac{d}{dz}(\sqrt{z}) \cdot \dfrac{dz}{dx}$ (by the chain rule)

$=\dfrac{1}{2\sqrt{z}} \cdot \dfrac{d}{dx}(\sin x)$ by Equation (i)

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

$=\dfrac{\cos x}{2\sqrt{\sin x}}$  $[\because z=\sin x]$

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