What is the derivative of (7x/(2^x)) ?

The derivative of (7x/(2^x)) is (7(1-ln(2)x))/(2^x)

What is the first derivative of (7x/(2^x)) ?

The first derivative of (7x/(2^x)) is (7(1-ln(2)x))/(2^x)

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

\frac{7 ( 1 - ln ( 2 ) x )}{2 ^{x}}

Solution steps

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

Take the constant out:

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

= 7 \frac{d}{d x} (\frac{x}{2 ^{x}})

Apply the Quotient Rule:

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

= 7 \cdot \frac{\frac{d x}{d x} 2 ^{x} - \frac{d}{d x} ( 2 ^{x} ) x}{( 2 ^{x} ) ^{2}}

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

\frac{d}{d x} (2^{x}) = 2^{x} ln (2)

= 7 \cdot \frac{1 \cdot 2 ^{x} - 2 ^{x} ln ( 2 ) x}{( 2 ^{x} ) ^{2}}

Simplify

7 \cdot \frac{1 \cdot 2 ^{x} - 2 ^{x} ln ( 2 ) x}{( 2 ^{x} ) ^{2}} : \frac{7 ( 1 - ln ( 2 ) x )}{2 ^{x}}

= \frac{7 ( 1 - ln ( 2 ) x )}{2 ^{x}}

\frac{d y}{d x} = 8 x^{7} - 2 sec^{2} (x)

\int \frac{( x ^{2} )}{( x ^{3} + 1 ) ^{2}} d x

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (2 t)

\int \frac{sec ( ln ( x ) ) tan ( ln ( x ) )}{x} d x

(- 2 x)^{^{'}}

What is the derivative of (7x/(2^x)) ?
The derivative of (7x/(2^x)) is (7(1-ln(2)x))/(2^x)
What is the first derivative of (7x/(2^x)) ?
The first derivative of (7x/(2^x)) is (7(1-ln(2)x))/(2^x)