What is the ((x-2)^2(x+4))^' ?

The ((x-2)^2(x+4))^' is 3x^2-12

What is the first ((x-2)^2(x+4))^' ?

The first ((x-2)^2(x+4))^' is 3x^2-12

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((x - 2)^{2} (x + 4))^{^{'}}

3 x^{2} - 12

Solution steps

((x - 2)^{2} (x + 4))^{'}

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = (x - 2)^{2}, g = x + 4

= ((x - 2)^{2})^{^{'}} (x + 4) + (x + 4)^{^{'}} (x - 2)^{2}

((x - 2)^{2})^{'} = 2 (x - 2)

(x + 4)^{'} = 1

= 2 (x - 2) (x + 4) + 1 \cdot (x - 2)^{2}

Simplify

2 (x - 2) (x + 4) + 1 \cdot (x - 2)^{2} : 3 x^{2} - 12

= 3 x^{2} - 12

n = 1 \sum \infty \frac{n}{3 ^{n^{2}}}

f (x) = 5 x^{2} e^{3 x}

y = (1 + 2 x)^{2}, (3, 49)

\frac{s}{( s + 1 ) ^{2} + 4}

\frac{\partial}{\partial y} (x^{2} + y^{2} + z^{2})