What is the derivative of (4e^{2x}-2e^x+1/(2e^{2x)}) ?

The derivative of (4e^{2x}-2e^x+1/(2e^{2x)}) is (e^x-1)/(e^{2x)}

What is the first derivative of (4e^{2x}-2e^x+1/(2e^{2x)}) ?

The first derivative of (4e^{2x}-2e^x+1/(2e^{2x)}) is (e^x-1)/(e^{2x)}

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\frac{d}{d x} (\frac{4 e ^{2 x} - 2 e ^{x} + 1}{2 e ^{2 x}})

\frac{e ^{x} - 1}{e ^{2 x}}

Solution steps

\frac{d}{d x} (\frac{4 e ^{2 x} - 2 e ^{x} + 1}{2 e ^{2 x}})

Take the constant out:

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

= \frac{1}{2} \frac{d}{d x} (\frac{4 e ^{2 x} - 2 e ^{x} + 1}{e ^{2 x}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (4 e^{2 x} - 2 e^{x} + 1) = - 2 e^{x} + 8 e^{2 x}

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

= \frac{1}{2} \cdot \frac{( - 2 e ^{x} + 8 e ^{2 x} ) e ^{2 x} - e ^{2 x} \cdot 2 ( 4 e ^{2 x} - 2 e ^{x} + 1 )}{( e ^{2 x} ) ^{2}}

Simplify

\frac{1}{2} \cdot \frac{( - 2 e ^{x} + 8 e ^{2 x} ) e ^{2 x} - e ^{2 x} \cdot 2 ( 4 e ^{2 x} - 2 e ^{x} + 1 )}{( e ^{2 x} ) ^{2}} : \frac{e ^{x} - 1}{e ^{2 x}}

= \frac{e ^{x} - 1}{e ^{2 x}}

\int 3 x^{2} e^{x} d x

\frac{\partial}{\partial x} (- 2 y csc^{3} (x))

\frac{d}{d x} (2 cos^{2} (x^{2} + 2))

\int \frac{3}{2} x e^{- 6 x} d x

6 + \frac{1}{5} sin (6 π t)

What is the derivative of (4e^{2x}-2e^x+1/(2e^{2x)}) ?
The derivative of (4e^{2x}-2e^x+1/(2e^{2x)}) is (e^x-1)/(e^{2x)}
What is the first derivative of (4e^{2x}-2e^x+1/(2e^{2x)}) ?
The first derivative of (4e^{2x}-2e^x+1/(2e^{2x)}) is (e^x-1)/(e^{2x)}