What is the derivative of (3x+3/(4e^x)) ?

The derivative of (3x+3/(4e^x)) is -(3x)/(4e^x)

What is the first derivative of (3x+3/(4e^x)) ?

The first derivative of (3x+3/(4e^x)) is -(3x)/(4e^x)

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\frac{d}{d x} (\frac{3 x + 3}{4 e ^{x}})

- \frac{3 x}{4 e ^{x}}

Solution steps

\frac{d}{d x} (\frac{3 x + 3}{4 e ^{x}})

Take the constant out:

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

= \frac{1}{4} \frac{d}{d x} (\frac{3 x + 3}{e ^{x}})

Apply the Quotient Rule:

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

= \frac{1}{4} \cdot \frac{\frac{d}{d x} ( 3 x + 3 ) e ^{x} - \frac{d}{d x} ( e ^{x} ) ( 3 x + 3 )}{( e ^{x} ) ^{2}}

\frac{d}{d x} (3 x + 3) = 3

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

= \frac{1}{4} \cdot \frac{3 e ^{x} - e ^{x} ( 3 x + 3 )}{( e ^{x} ) ^{2}}

Simplify

\frac{1}{4} \cdot \frac{3 e ^{x} - e ^{x} ( 3 x + 3 )}{( e ^{x} ) ^{2}} : - \frac{3 x}{4 e ^{x}}

= - \frac{3 x}{4 e ^{x}}

\int \frac{2 x + 1}{x + 2} d x

\frac{d y}{d x} = (y + 5) (x - 2)

\frac{d}{d x} (2^{s i n (x)} \cdot t a g (x, t) (x - x^{3}))

\int \frac{x - 1}{x ^{3} + x} d x

f (x) = x e^{4 x}

What is the derivative of (3x+3/(4e^x)) ?
The derivative of (3x+3/(4e^x)) is -(3x)/(4e^x)
What is the first derivative of (3x+3/(4e^x)) ?
The first derivative of (3x+3/(4e^x)) is -(3x)/(4e^x)