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

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

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

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

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

e^{4 x} + 4 e^{4 x} x

Solution steps

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

Apply the Product Rule:

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

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

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

x^{'} = 1

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

= 1 \cdot e^{4 x} + e^{4 x} \cdot 4 x

Simplify

= e^{4 x} + 4 e^{4 x} x

arcsin (x^{3})

f (x) = 7 - x^{3}

\frac{d}{d x} (\frac{5 x}{x ^{2} + 4})

x \frac{d y}{d x} + y = x^{2} y^{2}