What is the derivative of (5x^4)/(e^x) ?

The derivative of (5x^4)/(e^x) is (5x^3(4-x))/(e^x)

What is the first derivative of (5x^4)/(e^x) ?

The first derivative of (5x^4)/(e^x) is (5x^3(4-x))/(e^x)

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derivative of

\frac{5 x ^{4}}{e ^{x}}

\frac{5 x ^{3} ( 4 - x )}{e ^{x}}

Solution steps

\frac{d}{d x} (\frac{5 x ^{4}}{e ^{x}})

Take the constant out:

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

= 5 \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}}

= 5 \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}

= 5 \cdot \frac{4 x ^{3} e ^{x} - e ^{x} x ^{4}}{( e ^{x} ) ^{2}}

Simplify

5 \cdot \frac{4 x ^{3} e ^{x} - e ^{x} x ^{4}}{( e ^{x} ) ^{2}} : \frac{5 x ^{3} ( 4 - x )}{e ^{x}}

= \frac{5 x ^{3} ( 4 - x )}{e ^{x}}

\frac{\partial}{\partial y} (x - y^{2} + z^{2})

d er i v a t i v e 6 e^{2 t}

a re a y = 3 x^{3}, y = 75 x

d er i v a t i v e x - 3

d er i v a t i v e 7 π r^{2}

What is the derivative of (5x^4)/(e^x) ?
The derivative of (5x^4)/(e^x) is (5x^3(4-x))/(e^x)
What is the first derivative of (5x^4)/(e^x) ?
The first derivative of (5x^4)/(e^x) is (5x^3(4-x))/(e^x)