Question

find the derivative of $x^{2}(x - 1)^{5}$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u = x^{2}$ and $v=(x - 1)^{5}$. First, find $u^\prime$ and $v^\prime$.
$u^\prime=\frac{d}{dx}(x^{2}) = 2x$

Step2: Apply chain - rule to find $v^\prime$

The chain - rule states that if $y = f(g(x))$, then $y^\prime=f^\prime(g(x))\cdot g^\prime(x)$. For $v=(x - 1)^{5}$, let $g(x)=x - 1$ and $f(u)=u^{5}$. Then $g^\prime(x)=1$ and $f^\prime(u) = 5u^{4}$. So $v^\prime=5(x - 1)^{4}\cdot1=5(x - 1)^{4}$

Step3: Calculate the derivative of the product

Using the product - rule $y^\prime=u^\prime v+uv^\prime$, substitute $u = x^{2}$, $u^\prime = 2x$, $v=(x - 1)^{5}$, and $v^\prime=5(x - 1)^{4}$ into it.
$y^\prime=2x(x - 1)^{5}+x^{2}\cdot5(x - 1)^{4}$

Step4: Factor out common terms

Factor out $x(x - 1)^{4}$ from the above expression.
$y^\prime=x(x - 1)^{4}[2(x - 1)+5x]$

Step5: Simplify the expression inside the brackets

Expand and simplify $2(x - 1)+5x$.
$2(x - 1)+5x=2x-2 + 5x=7x-2$
So $y^\prime=x(x - 1)^{4}(7x - 2)$

Answer:

$x(x - 1)^{4}(7x - 2)$