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 e x d x

y^{^{''}} + 4 y^{^{'}} + 4 y = 0

\frac{d}{d x} (\frac{24}{( 1 - x ) ^{5}})

(ln ∣ x + 2 ∣)^{7}

x^{^{''}} = cos (t)

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)