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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

3 y^{2} = 2 y (x^{4} - x) \frac{d y}{d x} - 3

\int_{2}^{3} \frac{30}{3 - x} d x

\int e^{- t} \cdot t d t

y^{^{''}} + 7 y^{^{'}} + 12 y = e^{2 t} \cdot sin (3 t)

x \to - 5 lim (ln (x + 5))

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