Question

What is the derivative of \[f(x) = {(2x - 3)^4}{({x^2} + x + 1)^5}\] ?

Answer 1

Hint: We use Product – Rule to find the derivative of given function\[f(x) = {(2x - 3)^4}{({x^2} + x + 1)^5}\]. The Product rule helps us to differentiate between two or more of the functions in a given function.
If $u$ and $v$ are the two given function of $x$ then the product rule is given by the following formula:
$\dfrac{{d(uv)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$. We also know that $\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$.

Complete step by step solution:
The given function is the product of two functions ${(2x - 3)^4}$ and ${({x^2} + x + 1)^5}$.
We use product rule to find the derivative, that means first we multiply the first function by the derivative of the second function and the second function is multiplied by the derivative of the first function and add them.
Let us assume $u = {(2x - 3)^4}$ and $v = {({x^2} + x + 1)^5}$
We will use the formula $\dfrac{{d(uv)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
So,
${{(2x - 3)}^4} \dfrac{{d{{({x^2} + x + 1)}^5}}}{{dx}} = {(2x - 3)^4}\dfrac{{d({x^2} + x + 1)}}{{dx}} + ({x^2} + x + 1)\dfrac{{d{{(2x - 3)}^4}}}{{dx}}$$ \ldots \ldots (1)$
$ \Rightarrow \dfrac{{d{{({x^2} + x + 1)}^5}}}{{dx}} = 5{({x^2} + x + 1)^4}(2x + 1)$$ \ldots \ldots (2)$
$ \Rightarrow {(2x - 3)^4} = 4{(2x - 3)^3}(2)$$ \ldots \ldots (3)$
Substitute the value of equation $(2)$ and $(3)$ in equation $(1)$. We get,
$ \Rightarrow {(2x - 3)^4}5{({x^2} + x + 1)^4}(2x + 1) + {({x^2} + x + 1)^5}4{(2x - 3)^3}(2)$
Taking ${(2x - 3)^3}$ and ${({x^2} + x + 1)^4}$ common from the equation. We get,
$ \Rightarrow {(2x - 3)^3}{({x^2} + x + 1)^4}[5(2x + 1) + ({x^2} + x + 1)4(2)]$
Simplifying the above equation. We get,
$ \Rightarrow {(2x - 3)^3}{({x^2} + x + 1)^4}[10x + 5 + 8{x^2} + 8x + 8)$
$ \Rightarrow {(2x - 3)^3}{({x^2} + x + 1)^4}(8{x^2} + 18x + 13)$
Hence, the derivative of the function \[f(x) = {(2x - 3)^4}{({x^2} + x + 1)^5}\] is${(2x - 3)^3}{({x^2} + x + 1)^4}(8{x^2} + 18x + 13)$.

Note:
In this type of problem one should not forget to derivate the inner function. First calculate the derivative of the outer function and multiply it by the derivative of inner function. Use chain rule to solve these problems. Chain rule is used where the function is composite, we can denote chain rule by $f.g$, where $f$ and $g$ are two functions. Chain rule states that the derivative of a composite function can be taken as the derivative of the outer function which we multiply by the derivative of the inner function.