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

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

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

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

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derivative of

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

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

Solution steps

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

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

\frac{d x}{d x} = 1

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

Simplify

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

2 x + 3 y

\int_{- 2}^{- 1} (x^{2} - x - 2) d x

f (x) = e^{x} ln (x)

x \to 0 lim (\frac{sin ( 4 x )}{6})

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

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