Question

Derivative of $\dfrac{1}{\sqrt{x}}$ is
A. $\dfrac{1}{2x\sqrt{x}}$
B. $-\dfrac{1}{2x\sqrt{x}}$
C. $2x\sqrt{x}$
D. $-2x\sqrt{x}$

Answer 1

Here we have been given a variable and we have to find its derivative. We will use the basic formula that is $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ to solve the given values. Firstly we will bring the value from the denominator to the numerator by changing the sign of its power and then we will use the formula to get our desired answer.

Complete step by step answer:
We have to find the derivative of the value given below:
$\dfrac{1}{\sqrt{x}}$
We will rewrite the above value as follows:
$\Rightarrow \dfrac{1}{\sqrt{x}}=\dfrac{1}{{{x}^{\dfrac{1}{2}}}}$
We know ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$ so,
$\Rightarrow \dfrac{1}{\sqrt{x}}={{x}^{-\dfrac{1}{2}}}$…..$\left( 1 \right)$
Now we know the formula to find the derivative of a variable is as follows,
$\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$
On comparing the left side of the above value with equation (1) we get ${{x}^{n}}={{x}^{-\dfrac{1}{2}}}$ ,
So using the formula,
$\dfrac{d}{dx}\left( {{x}^{-\dfrac{1}{2}}} \right)=-\dfrac{1}{2}\times {{x}^{-\dfrac{1}{2}-1}}$
Now using the exponent property ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$ above we get,
$\dfrac{d}{dx}\left( {{x}^{-\dfrac{1}{2}}} \right)=-\dfrac{1}{2{{x}^{\dfrac{1}{2}+1}}}$
We know that ${{a}^{m+n}}={{a}^{m}}{{a}^{n}}$ using it above we get,
$\Rightarrow \dfrac{d}{dx}\left( {{x}^{-\dfrac{1}{2}}} \right)=-\dfrac{1}{2{{x}^{1}}{{x}^{\dfrac{1}{2}}}}$
$\Rightarrow \dfrac{d}{dx}\left( {{x}^{-\dfrac{1}{2}}} \right)=-\dfrac{1}{2x\sqrt{x}}$
Put the value from equation (1) above,
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{\sqrt{x}} \right)=-\dfrac{1}{2x\sqrt{x}}$
So we get the derivative of $\dfrac{1}{\sqrt{x}}$ as $-\dfrac{1}{2x\sqrt{x}}$ .
Hence the correct option is (B).

Note:
Derivative defines the rate of change of a function with respect to its independent variable. It is used to represent the amount by which the given function is changing at a certain point. In this question we have used the exponent property as well so that we have to do calculations. As the answers in the options were given in a simplified form so we changed the values accordingly and that is why we didn’t solve exponent power in the numerator itself. Differentiation is one of the important concepts of calculus apart from integration and differentiation is a method to find the derivative of the function.