Question

How do you find the derivative of $x{\left( {x - 4} \right)^3}$ ?

Answer 1

Hint: In order to find the derivative, we use the product rule of differentiation. Again we use the chain rule for finding some term and then putting the value in the given term. On doing some simplification we get the required answer.

Formula used: $\dfrac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\dfrac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right)\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$
$\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$

Complete step-by-step solution:
In the given question, we need to find the derivative of the expression $x{\left( {x - 4} \right)^3}$.
In order to do so, we use the product rule of differentiation.
Thus the formula for the product rule of differentiation is as follows: $\dfrac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\dfrac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right)\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$
According to the question: $f\left( x \right) = x$ and $g\left( x \right) = {\left( {x - 4} \right)^3}$
Thus, $\dfrac{d}{{dx}}\left( {x{{\left( {x - 4} \right)}^3}} \right) = \left( x \right)\dfrac{d}{{dx}}\left( {{{\left( {x - 4} \right)}^3}} \right) + {\left( {x - 4} \right)^3}\dfrac{d}{{dx}}\left( x \right)....\left( A \right)$.
Let us call this equation (A)
Now, we use the chain rule to differentiate ${\left( {x - 4} \right)^3}$
According to the chain rule, we know that $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$
Let $u = x - 4$
Therefore, differentiating both sides of the above expression with respect to $x$, we get:
\[ \Rightarrow \dfrac{{du}}{{dx}} = 1\]
Now, let us take$y = {u^3}$, and we differentiate both sides with respect to $u$, thus we get:
$ \Rightarrow \dfrac{{dy}}{{du}} = 3{u^2}$
Now, we know that $u = x - 4$, therefore $3{u^2} = 3{\left( {x - 4} \right)^2}$
Thus,$\dfrac{{dy}}{{dx}} = \left\{ {1 \times 3{{\left( {x - 4} \right)}^2}} \right\}$
On simplify we get,
$ \Rightarrow \dfrac{d}{{dx}}{\left( {x - 4} \right)^3} = 3{\left( {x - 4} \right)^2}$
Placing the derivative of ${\left( {x - 4} \right)^3}$ in equation (A), we get:
$ \Rightarrow \dfrac{d}{{dx}}\left( {x{{\left( {x - 4} \right)}^3}} \right) = x\left( {3{{\left( {x - 4} \right)}^2}} \right) + {\left( {x - 4} \right)^3}\dfrac{d}{{dx}}\left( x \right)$
Now, we know that $\dfrac{d}{{dx}}\left( x \right) = 1$
Therefore, $\dfrac{d}{{dx}}\left( {x{{\left( {x - 4} \right)}^3}} \right) = 3x{\left( {x - 4} \right)^2} + {\left( {x - 4} \right)^3}$
Now we simplify the given expression further. We take $\left( {x - 4} \right)$ as a common factor outside.
$ \Rightarrow \dfrac{d}{{dx}}\left( {x{{\left( {x - 4} \right)}^3}} \right) = {\left( {x - 4} \right)^2}{\left( {3x + x - 4} \right)^{}}$
On simplifying further, we get:
$ \Rightarrow \dfrac{d}{{dx}}\left( {x{{\left( {x - 4} \right)}^3}} \right) = {\left( {x - 4} \right)^2}\left( {4x - 4} \right)$

Therefore ${\left( {x - 4} \right)^2}\left( {4x - 4} \right)$ is the required answer.

Note: Derivatives are a fundamental tool of Calculus, which is a branch of Mathematics. Derivation or differentiation is a way of finding the instantaneous rate of change of a function based on one of its variables. We can find the derivative of trigonometric functions, logarithmic functions, etc using different formulas.
Some common formulas of differentiation are:
$\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$
$\dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x}$
$\dfrac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\dfrac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right)\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$