What is the (x^2e^{4x})^' ?

The (x^2e^{4x})^' is 2xe^{4x}+e^{4x}*4x^2

What is the first (x^2e^{4x})^' ?

The first (x^2e^{4x})^' is 2xe^{4x}+e^{4x}*4x^2

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(x^{2} e^{4 x})^{^{'}}

2 x e^{4 x} + e^{4 x} \cdot 4 x^{2}

Solution steps

(x^{2} e^{4 x})^{'}

Apply the Product Rule:

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

f = x^{2}, g = e^{4 x}

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

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

(e^{4 x})^{'} = e^{4 x} \cdot 4

= 2 x e^{4 x} + e^{4 x} \cdot 4 x^{2}

x \to 0 lim (\frac{x - 3}{2 x - 3})

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

x \to 4 lim (8 x^{2} - 9 x + 1)

\frac{\partial}{\partial x} (y^{0.5} x)

f (x) = x^{3} - 4 x^{2}, at x = 0

What is the (x^2e^{4x})^' ?
The (x^2e^{4x})^' is 2xe^{4x}+e^{4x}*4x^2
What is the first (x^2e^{4x})^' ?
The first (x^2e^{4x})^' is 2xe^{4x}+e^{4x}*4x^2