
How do you find the derivative of $\dfrac{4}{{\sqrt x }}$?
Answer
545.1k+ views
Hint: We are given an expression and we have to find its derivative. When we are differentiating things involving radicals it's never a bad idea to rewrite the radicals as rational exponents. After it becomes easy to differentiate using the power rule i.e.
$\dfrac{d}{{dx}}a{x^n} = an{x^{n - 1}}$
To use the power rule, multiply the variable’s exponent n, by its coefficient a, then subtract 1 from the exponent. If there is no coefficient (the coefficient is 1), then the exponent will become the new coefficient.
Here $x$ is any variable and $n$ is the exponent. By using this we will find the derivative of the expression.
Complete step-by-step answer:
Step 1: We are given a mathematical expression i.e. $\dfrac{4}{{\sqrt x }}$ and we have to find its derivative. Firstly we will rewrite the expression in the radical form i.e. rational number having an exponent.
$\dfrac{4}{{\sqrt x }} = 4x^{-1/2}$
$\Rightarrow y= 4x^{-1/2}$
Step 2: Now, this is easily differentiated using the power rule:
$\dfrac{d}{{dx}}a{x^n} = an{x^{n - 1}}$
Here, $a = 4;n = - \dfrac{1}{2};x = x$
Using this we will differentiate:
$\dfrac{dy}{dx}= \dfrac{-1}{2}.4x^{-3/2}$
Now we will simplify it:
$\dfrac{dy}{dx}= -2x^{-3/2}$
And we can convert it in this form also
$\dfrac{{dy}}{{dx}} = - \dfrac{2}{{\sqrt {{x^3}} }}$
Hence the derivative is $ - \dfrac{2}{{\sqrt {{x^3}} }}$
Note:
This type of question has an expression which can be converted into radical form then just use a power rule and solve it. There is no easier way than this sometime students try to apply first principles to find the derivative but unless it's mentioned in the question to solve by first principle don't use it because it makes a question lengthy and sometimes so messed up. Students mainly make mistakes only. In these questions, use a simple power rule which is short and easy.
Commit to memory:
$\dfrac{d}{{dx}}a{x^n} = an{x^{n - 1}}$
$\dfrac{d}{{dx}}a{x^n} = an{x^{n - 1}}$
To use the power rule, multiply the variable’s exponent n, by its coefficient a, then subtract 1 from the exponent. If there is no coefficient (the coefficient is 1), then the exponent will become the new coefficient.
Here $x$ is any variable and $n$ is the exponent. By using this we will find the derivative of the expression.
Complete step-by-step answer:
Step 1: We are given a mathematical expression i.e. $\dfrac{4}{{\sqrt x }}$ and we have to find its derivative. Firstly we will rewrite the expression in the radical form i.e. rational number having an exponent.
$\dfrac{4}{{\sqrt x }} = 4x^{-1/2}$
$\Rightarrow y= 4x^{-1/2}$
Step 2: Now, this is easily differentiated using the power rule:
$\dfrac{d}{{dx}}a{x^n} = an{x^{n - 1}}$
Here, $a = 4;n = - \dfrac{1}{2};x = x$
Using this we will differentiate:
$\dfrac{dy}{dx}= \dfrac{-1}{2}.4x^{-3/2}$
Now we will simplify it:
$\dfrac{dy}{dx}= -2x^{-3/2}$
And we can convert it in this form also
$\dfrac{{dy}}{{dx}} = - \dfrac{2}{{\sqrt {{x^3}} }}$
Hence the derivative is $ - \dfrac{2}{{\sqrt {{x^3}} }}$
Note:
This type of question has an expression which can be converted into radical form then just use a power rule and solve it. There is no easier way than this sometime students try to apply first principles to find the derivative but unless it's mentioned in the question to solve by first principle don't use it because it makes a question lengthy and sometimes so messed up. Students mainly make mistakes only. In these questions, use a simple power rule which is short and easy.
Commit to memory:
$\dfrac{d}{{dx}}a{x^n} = an{x^{n - 1}}$
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

