Question

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

Answer 1

Hint: To differentiate the function $Y = {x^2}\left( {2x + 3} \right)$ use product rule. Product rule states that if there are two functions say \[f\left( x \right)\] and \[g\left( x \right)\] then the derivative of \[f\left( x \right) \cdot g\left( x \right)\] will be \[\dfrac{d}{{dx}}\left( {f\left( x \right) \cdot g\left( x \right)} \right) = f\left( x \right)\dfrac{{d\left( {g\left( x \right)} \right)}}{{dx}} + g\left( x \right)\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}}\].

Complete step by step solution:
We have the given function $Y = {x^2}\left( {2x + 3} \right).........\left( i \right)$.
Apply the product rule on the given function $Y = {x^2}\left( {2x + 3} \right)$ while finding its derivative. So, since \[{x^2}\] is the first function and \[2x + 3\] is the second function then let \[f\left( x \right) = {x^2}\] and \[g\left( x \right) = 2x + 3\] hence \[Y = f\left( x \right) \cdot g\left( x \right)\].
Differentiating equation (i) we get,
\[ \Rightarrow \dfrac{{dY}}{{dx}} = f\left( x \right)\dfrac{{d\left( {g\left( x \right)} \right)}}{{dx}} + g\left( x \right)\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}}\]
On putting values \[f\left( x \right) = {x^2}\] and \[g\left( x \right) = 2x + 3\]
\[ \Rightarrow \dfrac{{dY}}{{dx}} = {x^2}\dfrac{d}{{dx}}\left( {2x + 3} \right) + \left( {2x + 3} \right)\dfrac{d}{{dx}}\left( {{x^2}} \right)......\left( {ii} \right)\]
Differentiate the \[{x^2}\] and \[2x + 3\] in the equation (ii).
\[ \Rightarrow \dfrac{{dY}}{{dx}} = {x^2} \cdot \left( {2\dfrac{{dx}}{{dx}} + \dfrac{{d3}}{{dx}}} \right) + \left( {2x + 3} \right) \cdot \dfrac{{d{x^2}}}{{dx}}\]
And by power rule, we get
\[ \Rightarrow \dfrac{{dY}}{{dx}} = {x^2} \cdot \left( {2 + 0} \right) + \left( {2x + 3} \right) \cdot \left( {2x} \right)\]
\[ \Rightarrow \dfrac{{dY}}{{dx}} = 2{x^2} + 4{x^2} + 6x........\left( {iii} \right)\]
Now we have got the equation (iii). In this equation add \[4{x^2}\] with \[2{x^2}\].
\[ \Rightarrow \dfrac{{dY}}{{dx}} = 2{x^2} + 4{x^2} + 6x\]
\[ \Rightarrow \dfrac{{dY}}{{dx}} = 6{x^2} + 6x.....\left( {iv} \right)\]
Make the factor in the equation (iv). Take \[6x\] common from the equation and then write the resulting equation.
\[ \Rightarrow \dfrac{{dY}}{{dx}} = 6{x^2} + 6x\]
\[ \Rightarrow \dfrac{{dY}}{{dx}} = 6x\left( {x + 1} \right)\]
Thus, we have obtained the differential of the given function $Y = {x^2}\left( {2x + 3} \right)$ as \[\dfrac{{dY}}{{dx}} = 6x\left( {x + 1} \right)\].
Hence, the derivative of the function $Y = {x^2}\left( {2x + 3} \right)$ is \[6x\left( {x + 1} \right)\].

Note:
While differentiating a function we should always keep in mind that we know the formula for differentiation i.e. \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]. If two functions are in product form then we should never forget to apply the product rule of differentiation. One alternate way of solving the differentiation of the given function is that we multiply the entire element outside of the bracket to the elements on the inside or in other words opening the bracket and then differentiate them altogether.
The alternate way will be:
\[\begin{array}{c}
\dfrac{{dY}}{{dx}} = {x^2}\left( {2x + 3} \right)\\
 = \dfrac{d}{{dx}}\left( {2{x^3} + 3{x^2}} \right)\\
 = 2\dfrac{{d\left( {{x^3}} \right)}}{{dx}} + 3\dfrac{{d\left( {{x^2}} \right)}}{{dx}}\\
 = 2 \cdot 3{x^2} + 3 \cdot 2x\\
 = 6{x^2} + 6x\\
 = 6x\left( {x + 1} \right)
\end{array}\]
In the alternate method the application of product rule is not required. This method is more simpler than the product rule method but this method would not work if there is any other variable than \[x\].