What is the first (3xe^x)^' ?

The first (3xe^x)^' is 3(e^x+e^xx)

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(3xe^x)^'

(3 x e^{x})^{^{'}}

3 (e^{x} + e^{x} x)

Solution steps

(3 x e^{x})^{'}

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 (x e^{x})^{^{'}}

Apply the Product Rule:

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

f = x, g = e^{x}

= 3 (x^{^{'}} e^{x} + (e^{x})^{^{'}} x)

x^{'} = 1

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

= 3 (1 \cdot e^{x} + e^{x} x)

Simplify

= 3 (e^{x} + e^{x} x)

\int x \frac{d}{1} (y) d x

\frac{\partial}{\partial x} (x \cdot ln (x + 2 y) + \frac{x}{y})

x \to 3 - lim (\frac{x - 3}{∣ x - 3 ∣})

(- 7) (0.3)

(2, 0), (- 2.5, 1)