Question

Find the derivative of the following: $(5{x^3} + 3x - 1)\left( {x - 1} \right)$.

Answer 1

Differentiation is nothing but finding the derivative. Let us collect the formulae that we will be using for this problem.
Let $k$be constant and $x$be a variable. $\dfrac{{dk}}{{dx}} = 0$, $\dfrac{{dx}}{{dx}} = 1$, $\dfrac{{dkx}}{{dx}} = 1$and $\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$.
These are some basic differentiation formulae that we need to know.
Formula used:
When there is a combination of two function which can be differentiated, we use a formula to differentiate it, that is$\dfrac{d}{{dx}}(uv) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$.

Complete step by step answer:
The given expression is$(5{x^3} + 3x - 1)\left( {x - 1} \right)$. Let us give a name for this expression,
$f(x) = (5{x^3} + 3x - 1)\left( {x - 1} \right)$
Our aim is to find the derivative of this function. We can see that the above function is in the form$f(x) = uv$. So, we have to use the formula $\dfrac{d}{{dx}}(uv) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$ to differentiate it.
Derivative of the function $f(x)$can be denoted by$f'(x)$.
$\therefore f'(x) = (5{x^3} + 3x - 1)\dfrac{d}{{dx}}(x - 1) + (x - 1)\dfrac{d}{{dx}}(5{x^3} + 3x - 1)$
$\Rightarrow f'(x) = (5{x^3} + 3x - 1)(1 - 0) + (x - 1)(5(3{x^2}) + 3 - 0)$
Simplifying the above equation, we get
$\Rightarrow f'(x) = (5{x^3} + 3x - 1) + (x - 1)(15{x^2} + 3)$
Let us multiply the factors to get a single expression.
$\Rightarrow f'(x) = (5{x^3} + 3x - 1) + (15{x^3} + 3x - 15{x^2} - 3)$
Now let’s group the like terms,
$\Rightarrow f'(x) = \left( {5{x^3} + 15{x^3}} \right) + \left( {3x + 3x} \right) - 15{x^2} - 3 - 1$
Simplify the like terms.
$\Rightarrow f'(x) = 20{x^3} + 6x - 15{x^2} - 4$
Rearranging the terms in the order,
$\Rightarrow f'(x) = 20{x^3} - 15{x^2} + 6x - 4$
Thus, this is the derivative of the given function.

Note: This problem can also be solved without using the formula$\dfrac{d}{{dx}}(uv) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$. First, we need to multiply the factors and make it a single expression and then it will be easier to differentiate. We can differentiate it term by term. This method is also easy as the method that we used to solve the problem. Basic derivative formulae are enough to solve this function since; there is no complexity in the terms. When there are trigonometry functions as terms in the equation we need to use derivatives of those functions.