What is the derivative of (40x^5)/(e^x) ?

The derivative of (40x^5)/(e^x) is (40x^4(5-x))/(e^x)

What is the first derivative of (40x^5)/(e^x) ?

The first derivative of (40x^5)/(e^x) is (40x^4(5-x))/(e^x)

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derivative of

\frac{40 x ^{5}}{e ^{x}}

\frac{40 x ^{4} ( 5 - x )}{e ^{x}}

Solution steps

\frac{d}{d x} (\frac{40 x ^{5}}{e ^{x}})

Take the constant out:

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

= 40 \frac{d}{d x} (\frac{x ^{5}}{e ^{x}})

Apply the Quotient Rule:

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

= 40 \cdot \frac{\frac{d}{d x} ( x ^{5} ) e ^{x} - \frac{d}{d x} ( e ^{x} ) x ^{5}}{( e ^{x} ) ^{2}}

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

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

= 40 \cdot \frac{5 x ^{4} e ^{x} - e ^{x} x ^{5}}{( e ^{x} ) ^{2}}

Simplify

40 \cdot \frac{5 x ^{4} e ^{x} - e ^{x} x ^{5}}{( e ^{x} ) ^{2}} : \frac{40 x ^{4} ( 5 - x )}{e ^{x}}

= \frac{40 x ^{4} ( 5 - x )}{e ^{x}}

\frac{d}{d x} (\frac{2}{x - 7})

(4 x y + 3 y^{2} - x) d x + (x^{2} + 2 x y) d y = 0

y = \frac{x ^{2}}{x + 4}, (3, \frac{9}{7})

\int_{- 7}^{7} (49 - x^{2})^{\frac{1}{2}} d x

y^{^{'}} + x y = 6 x y^{2}

What is the derivative of (40x^5)/(e^x) ?
The derivative of (40x^5)/(e^x) is (40x^4(5-x))/(e^x)
What is the first derivative of (40x^5)/(e^x) ?
The first derivative of (40x^5)/(e^x) is (40x^4(5-x))/(e^x)