What is the derivative of y=(9x)/(e^x) ?

The derivative of y=(9x)/(e^x) is (9(1-x))/(e^x)

What is the first derivative of y=(9x)/(e^x) ?

The first derivative of y=(9x)/(e^x) is (9(1-x))/(e^x)

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derivative of

y = \frac{9 x}{e ^{x}}

\frac{9 ( 1 - x )}{e ^{x}}

Solution steps

\frac{d}{d x} (\frac{9 x}{e ^{x}})

Take the constant out:

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

= 9 \frac{d}{d x} (\frac{x}{e ^{x}})

Apply the Quotient Rule:

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

= 9 \cdot \frac{\frac{d x}{d x} e ^{x} - \frac{d}{d x} ( e ^{x} ) x}{( e ^{x} ) ^{2}}

\frac{d x}{d x} = 1

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

= 9 \cdot \frac{1 \cdot e ^{x} - e ^{x} x}{( e ^{x} ) ^{2}}

Simplify

9 \cdot \frac{1 \cdot e ^{x} - e ^{x} x}{( e ^{x} ) ^{2}} : \frac{9 ( 1 - x )}{e ^{x}}

= \frac{9 ( 1 - x )}{e ^{x}}

f (x) = \frac{x + 1}{x ^{2} + 1}

\frac{d y}{d x} - 3 y = 2

\int \frac{1}{( x ^{2} - 9 )} d x

\int 3 (3 x + 4) 5 d x

\frac{d y}{d x} y = (2 x + 8)^{3}

What is the derivative of y=(9x)/(e^x) ?
The derivative of y=(9x)/(e^x) is (9(1-x))/(e^x)
What is the first derivative of y=(9x)/(e^x) ?
The first derivative of y=(9x)/(e^x) is (9(1-x))/(e^x)