What is the -derivative of 2^x ?

The -derivative of 2^x is -x derivative of 2^x+ln(2)*2^xx-2^x+C

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Find the Derivative - d/dx 2^x

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

- \frac{d}{d x} (2 sec^{2} (x) tan (x))

- \frac{d}{d x} (\frac{6}{x})

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

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

\int (e^{2 x} x^{2}) d x

What is the -derivative of 2^x ?
The -derivative of 2^x is -x derivative of 2^x+ln(2)*2^xx-2^x+C