Question

How do you find the derivative of ${\left( {2x - 1} \right)^3}$?

Answer 1

Hint: In the given question we need to differentiate the given function. Firstly, we observe the nature of the given function and which rule of differentiation can be applied to it. For the above function we can apply chain rule of differentiation which is given by,
$\dfrac{d}{{dx}}\left[ {{{\left( {f(x)} \right)}^n}} \right] = n{\left( {f(x)} \right)^{n - 1}} \cdot \dfrac{d}{{dx}}f(x)$
So apply the chain rule for the above function to solve and obtain the required derivative.

Complete step by step solution:
Given the function of the form ${\left( {2x - 1} \right)^3}$ …… (1)
We are asked to find the derivative for the function given in the equation (1).
In this problem, we use the chain rule of differentiation to find the derivative.
The chain rule of differentiation is given by,
$\dfrac{d}{{dx}}\left[ {{{\left( {f(x)} \right)}^n}} \right] = n{\left( {f(x)} \right)^{n - 1}} \cdot \dfrac{d}{{dx}}f(x)$
Note the in the above question $f(x) = 2x - 1$ and $n = 3$.
Now applying the chain rule we get,
$\dfrac{d}{{dx}}\left[ {{{\left( {2x - 1} \right)}^3}} \right] = 3 \cdot {\left( {2x - 1} \right)^{3 - 1}} \cdot \dfrac{d}{{dx}}(2x - 1)$ …… (2)
So now we find out the derivative of the function $2x - 1$. i.e. to find out the $\dfrac{d}{{dx}}(2x - 1)$.
$\dfrac{d}{{dx}}(2x - 1) = \dfrac{d}{{dx}}(2x) - \dfrac{d}{{dx}}(1)$
We know that the derivative of a constant term is zero. Hence, $\dfrac{d}{{dx}}(1) = 0$.
And $\dfrac{d}{{dx}}(2x) = 2$
Hence we get,
$ \Rightarrow \dfrac{d}{{dx}}(2x - 1) = \dfrac{d}{{dx}}(2x) - 0$
$ \Rightarrow \dfrac{d}{{dx}}(2x - 1) = 2$
Therefore the equation (2) becomes,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {2x - 1} \right)}^3}} \right] = 3 \cdot {\left( {2x - 1} \right)^{3 - 1}} \cdot (2)$
Simplifying this we get,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {2x - 1} \right)}^3}} \right] = 3 \cdot {\left( {2x - 1} \right)^2} \cdot (2)$
$ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {2x - 1} \right)}^3}} \right] = 6{\left( {2x - 1} \right)^2}$

Hence the derivative of the function ${\left( {2x - 1} \right)^3}$ is given by $6{\left( {2x - 1} \right)^2}$.

Note: Differentiation is a process of finding the derivative of a function. It is used to find the rates of change. The concept of differentiation also allows us to find the rate of change of the variable x with respect to the variable y.
Students must know the chain rule of differentiation and how to apply it to a given problem.
It is important to know which rule of differentiation can be applied, so that the given problem can be simplified.
The chain rule of differentiation is given by,
(1) $\dfrac{d}{{dx}}\left[ {{{\left( {f(x)} \right)}^n}} \right] = n{\left( {f(x)} \right)^{n - 1}} \cdot f'(x)$
(2) $\dfrac{d}{{dx}}\left[ {f\left( {g(x)} \right)} \right] = f'\left( {g(x)} \right) \cdot g'(x)$