What is the derivative of f(x)=(x(2-ln(2)x))/(2^x) ?

The derivative of f(x)=(x(2-ln(2)x))/(2^x) is (2-4ln(2)x+ln^2(2)x^2)/(2^x)

What is the first derivative of f(x)=(x(2-ln(2)x))/(2^x) ?

The first derivative of f(x)=(x(2-ln(2)x))/(2^x) is (2-4ln(2)x+ln^2(2)x^2)/(2^x)

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derivative of

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

\frac{2 - 4 ln ( 2 ) x + ln ^{2} ( 2 ) x ^{2}}{2 ^{x}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( 2 - 2 ln ( 2 ) x ) \cdot 2 ^{x} - 2 ^{x} ln ( 2 ) x ( 2 - ln ( 2 ) x )}{( 2 ^{x} ) ^{2}} : \frac{2 - 4 ln ( 2 ) x + ln ^{2} ( 2 ) x ^{2}}{2 ^{x}}

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

\frac{d}{d x} (\frac{x + 1}{4})

x \to 0 lim (\frac{sin ( 4 x )}{x} + 2)

\frac{\partial}{\partial x} (x^{3} + y^{2} - 5 x y)

4 y^{^{''}} - y^{^{'}} = 0

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = 0

What is the derivative of f(x)=(x(2-ln(2)x))/(2^x) ?
The derivative of f(x)=(x(2-ln(2)x))/(2^x) is (2-4ln(2)x+ln^2(2)x^2)/(2^x)
What is the first derivative of f(x)=(x(2-ln(2)x))/(2^x) ?
The first derivative of f(x)=(x(2-ln(2)x))/(2^x) is (2-4ln(2)x+ln^2(2)x^2)/(2^x)