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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

y^{^{'}} (x) = \frac{6 x ^{2}}{6 y ^{2} - 7}, y (1) = 0

e^{x} \frac{d y}{d x} + 3 e^{x} y = 2

y = x^{5^{x}}

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

\int \frac{1}{64 e ^{- 5 x} + e ^{5 x}} d x

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