What is the derivative of (2^x/x+x-log_{2}(x)) ?

The derivative of (2^x/x+x-log_{2}(x)) is (ln(2)*2^xx-2^x)/(x^2)+1-1/(xln(2))

What is the first derivative of (2^x/x+x-log_{2}(x)) ?

The first derivative of (2^x/x+x-log_{2}(x)) is (ln(2)*2^xx-2^x)/(x^2)+1-1/(xln(2))

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\frac{d}{d x} (\frac{2 ^{x}}{x} + x - lo g_{2} (x))

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

Solution steps

\frac{d}{d x} (\frac{2 ^{x}}{x} + x - lo g_{2} (x))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{2 ^{x}}{x}) + \frac{d x}{d x} - \frac{d}{d x} (lo g_{2} (x))

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

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

\frac{d}{d x} (lo g_{2} (x)) = \frac{1}{x ln ( 2 )}

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

What is the derivative of (2^x/x+x-log_{2}(x)) ?
The derivative of (2^x/x+x-log_{2}(x)) is (ln(2)*2^xx-2^x)/(x^2)+1-1/(xln(2))
What is the first derivative of (2^x/x+x-log_{2}(x)) ?
The first derivative of (2^x/x+x-log_{2}(x)) is (ln(2)*2^xx-2^x)/(x^2)+1-1/(xln(2))