What is the derivative of-1/(sqrt(2^x)) ?

The derivative of-1/(sqrt(2^x)) is ln(2)*2^{(-x-2)/2}

What is the first derivative of-1/(sqrt(2^x)) ?

The first derivative of-1/(sqrt(2^x)) is ln(2)*2^{(-x-2)/2}

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

= - \frac{1}{2 ( 2 ^{x} ) ^{\frac{3}{2}}} \frac{d}{d x} (2^{x})

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

= - (- \frac{1}{2 ( 2 ^{x} ) ^{\frac{3}{2}}} \cdot 2^{x} ln (2))

Simplify

- (- \frac{1}{2 ( 2 ^{x} ) ^{\frac{3}{2}}} \cdot 2^{x} ln (2)) : ln (2) \cdot 2^{\frac{- x - 2}{2}}

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

\frac{\partial}{\partial x} (1 0^{x})

\frac{x ^{2} - 2 x + 3}{x + 1}

\int \frac{1}{7} d x

\int sin (2 x) ln (sin (2 x)) d x

\frac{10}{s ( s ^{2} + 2 s + 10 )}

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