For j(x)=(1−x^2)(−3x^2+2x) find j′(0) by applying the product rule

Question

please reply with answer.

Answer 1

Product Rule: (d/dx)[f(x) * g(x)] = f(x)g'(x) + f'(x)g(x)

j(x) = (1 − x²)(−3x² + 2x)

j'(x) = (1 - x²)(-6x + 2) + (-2x)(-3x² + 2x)

j'(0) = (1 - 0²)(-6(0) + 2) + (-2(0))(-3(0)² + 2(0)) = 1(2) + 0(0) = 2

Answer 2

Hi,

product rule=

derivative of first * second + derivative of second * first

j(x)=(1−x^2)(−3x^2+2x)

j'(x)=(1−x^2)'(−3x^2+2x) +(1−x^2)(−3x^2+2x)'

j'(x)= -2x (−3x^2+2x) + ( -6x + 2 ) ( 1−x^2)

next step is simplify

j'(x)=

6x^3 - 4x -6x +2 + 6x^3 -2x^2= 12x^3 -2x^2-10x + 2

I hope it is useful,

Minoo