What is the derivative of f(x)=(2x)/(2e^{9x)} ?

The derivative of f(x)=(2x)/(2e^{9x)} is (1-9x)/(e^{9x)}

What is the first derivative of f(x)=(2x)/(2e^{9x)} ?

The first derivative of f(x)=(2x)/(2e^{9x)} is (1-9x)/(e^{9x)}

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derivative of

f (x) = \frac{2 x}{2 e ^{9 x}}

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (ln (3 x - 5))

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

x^{2} + y^{2}

\int_{0}^{1} 2 π (y) (y - y^{3}) d y

\int y^{6} (cos (3 y)) d y

What is the derivative of f(x)=(2x)/(2e^{9x)} ?
The derivative of f(x)=(2x)/(2e^{9x)} is (1-9x)/(e^{9x)}
What is the first derivative of f(x)=(2x)/(2e^{9x)} ?
The first derivative of f(x)=(2x)/(2e^{9x)} is (1-9x)/(e^{9x)}