What is the derivative of ye^{2x} ?

The derivative of ye^{2x} is (dy)/(dx)e^{2x}+e^{2x}*2y

What is the first derivative of ye^{2x} ?

The first derivative of ye^{2x} is (dy)/(dx)e^{2x}+e^{2x}*2y

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\frac{d}{d x} (y e^{2 x})

\frac{d y}{d x} e^{2 x} + e^{2 x} \cdot 2 y

Solution steps

\frac{d}{d x} (y e^{2 x})

Apply the Product Rule:

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

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

= \frac{d y}{d x} e^{2 x} + \frac{d}{d x} (e^{2 x}) y

\frac{d}{d x} (e^{2 x}) = e^{2 x} \cdot 2

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

f (x) = \frac{x}{7}

x \to 0 lim (\frac{sin ^{4} ( 2 x )}{4 x ^{4}})

\frac{d v}{d t} = 0.64 - \frac{v}{2940} \cdot 8, v (0) = 0

\int (t^{3} - 3 t) (2 t) d t

\frac{d x}{d t} = 8 x t^{5}

What is the derivative of ye^{2x} ?
The derivative of ye^{2x} is (dy)/(dx)e^{2x}+e^{2x}*2y
What is the first derivative of ye^{2x} ?
The first derivative of ye^{2x} is (dy)/(dx)e^{2x}+e^{2x}*2y