Question

How do you find the derivative of $y = \dfrac{5}{{2{x^2}}}?$

Answer 1

Hint : This question describes the operation of addition/ subtraction/ multiplication/ division. Also, we need to know the formula to differentiate the term ${x^n}$ . Also, we need to know the multiplication process between the fraction term and the whole number term. We need to know how we can convert the term ${x^{ - n}}$ into ${x^n}$ .

Complete step-by-step answer :
In this question, we would find the value of $\dfrac{{dy}}{{dx}}$ ,
Here $y$ is equal to $\dfrac{5}{{2{x^2}}}$
That is,
$y = \dfrac{5}{{2{x^2}}} \to \left( 1 \right)$
We know that the formula for differentiating the term ${x^n}$ is
$\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}} \to \left( 2 \right)$
If we have to find the derivative of ${x^{ - n}}$ , then the equation $\left( 2 \right)$ becomes,
$\dfrac{{d\left( {{x^{ - n}}} \right)}}{{dx}} = - n{x^{ - n - 1}} \to \left( 3 \right)$
Let’s solve the equation $\left( 1 \right)$ ,
$\left( 1 \right) \to y = \dfrac{5}{{2{x^2}}}$
We would find the value of $\dfrac{{dy}}{{dx}}$ , so the above equation can also be written as
$\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {\dfrac{5}{{2{x^2}}}} \right)}}{{dx}}$
The above equation can also be written as,
$\dfrac{{d\left( {\dfrac{5}{{2{x^2}}}} \right)}}{{dx}} = \dfrac{{d\left( {\dfrac{{5{x^{ - 2}}}}{2}} \right)}}{{dx}} \to \left( 4 \right)$
If the term ${x^2}$ is present in the denominator, it converts into ${x^{ - 2}}$ when we move the ${x^2}$ to numerator as shown in the above equation.
Let’s compare the equation $\left( 3 \right)$ and $\left( 4 \right)$ ,
$\left( 3 \right) \to \dfrac{{d\left( {{x^{ - n}}} \right)}}{{dx}} = - n{x^{ - n - 1}}$
$\left( 4 \right) \to \dfrac{{d\left( {\dfrac{5}{{2{x^2}}}} \right)}}{{dx}} = \dfrac{{d\left( {\dfrac{{5{x^{ - 2}}}}{2}} \right)}}{{dx}}$
So, we have
$\dfrac{{d\left( {{x^{ - 2}}} \right)}}{{dx}} = - 2{x^{ - 2 - 1}}$
$\dfrac{{d\left( {{x^{ - 2}}} \right)}}{{dx}} = - 2{x^{ - 3}} \to \left( 5 \right)$
So, the equation $\left( 4 \right)$ can also be written as,
$\left( 4 \right) \to \dfrac{{d\left( {\dfrac{5}{{2{x^2}}}} \right)}}{{dx}} = \dfrac{{d\left( {\dfrac{{5{x^{ - 2}}}}{2}} \right)}}{{dx}}$
$\dfrac{{d\left( {\dfrac{5}{{2{x^2}}}} \right)}}{{dx}} = \dfrac{5}{2}\dfrac{{d\left( {{x^{ - 2}}} \right)}}{{dx}} \to \left( 6 \right)$
We don’t need to find the derivative of constant terms. So, we can take outside the constant term from the derivative term.
Let’s substitute the equation $\left( 5 \right)$ in the equation $\left( 6 \right)$ , we get
$\left( 6 \right) \to \dfrac{{d\left( {\dfrac{5}{{2{x^2}}}} \right)}}{{dx}} = \dfrac{5}{2}\dfrac{{d\left( {{x^{ - 2}}} \right)}}{{dx}}$
$\dfrac{{d\left( {\dfrac{5}{{2{x^2}}}} \right)}}{{dx}} = \dfrac{5}{2}\left( { - 2{x^{ - 3}}} \right)$
By solving the above equation we get,
$\dfrac{{d\left( {\dfrac{5}{{2{x^2}}}} \right)}}{{dx}} = - 5\left( {{x^{ - 3}}} \right)$
The above equation can also be written as
$\dfrac{{d\left( {\dfrac{5}{{2{x^2}}}} \right)}}{{dx}} = \dfrac{{ - 5}}{{{x^3}}}$
So, the final answer is,
The derivative of $\dfrac{5}{{2{x^2}}}$ is $\dfrac{{ - 5}}{{{x^3}}}$ .
So, the correct answer is “ $\dfrac{{ - 5}}{{{x^3}}}$ ”.

Note : If the ${x^n}$ is present in the denominator it converts into ${x^{ - n}}$ when we move the ${x^n}$ into the numerator. We don’t need to find the derivation of the constant terms. We can take outside the constant term from the derivative term. It also describes the operation of addition/ subtraction/ multiplication/ division