Question

(x-1)^2 (x+1)^3

Answer 1

f(x) = $(x-1)^2(x+1)^3$

$d/dx (fx) = f'(x)$

inserting the function f(x) in the derivative equation .

$d/dx (x-1)^2(x+1)^3$

apply the product rule to the equation to open the parts ,the equation becomes

$d/dx(x-1)^2.(x+1)^3+(x-1)^2.d/dx(x+1)^3$

the equation contains powers and products, applying the power rule and chain rule for the inner functions.

$2(x-1).d/dx[x-1].(x+1)^3+(x-1)^2.3(x+1)^2.d/dx[x+1]$

since the differentiation is linear ,the summands can be calculated separately as.

$2(x-1)(d/dx[x]+d/dx[-1].(x+1)^3+(x-1)^2.3(x+1)^2.(d/dx[x] +d/dx[1]$

the value of d/dx [x] = 1 and d/dx(-1) = 0 also d/dx(1) = 0

replacing the values in the equation the equation becomes

$2(x-1)(1+0)(x+1)^3 + (x-1)^2.3(x+1)^2.(1+0)$

$2(x-1)(x+1)^3 + 3(x-1)^2.(x+1)^2$

the equation can be simplified as follows by factoring

$(x-1)(x+1)^2(5x-1)$

to find roots of the equation at 0

(x-1) =0 ; x =1

(x+1=0 ; x = -1

5x-1 = 0 : x = 1/5