What is (\partial)/(\partial x)(2^{-x}) ?

The answer to (\partial)/(\partial x)(2^{-x}) is -ln(2)*2^{-x}

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(\partial)/(\partial x)(2^{-x})

\frac{\partial}{\partial x} (2^{- x})

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

Solution steps

\frac{\partial}{\partial x} (2^{- x})

Apply exponent rule:

a^{b} = e^{b l n (a)}

2^{- x} = e^{(- x) l n (2)}

= \frac{\partial}{\partial x} (e^{(- x) l n (2)})

Apply the chain rule:

e^{(- x) l n (2)} \frac{\partial}{\partial x} ((- x) ln (2))

= e^{(- x) l n (2)} \frac{\partial}{\partial x} ((- x) ln (2))

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

= e^{(- x) l n (2)} (- ln (2))

Simplify

e^{(- x) l n (2)} (- ln (2)) : - ln (2) \cdot 2^{- x}

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

\int \frac{8}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

\int_{- 2}^{1} x^{3} d x

\frac{2}{t ^{2}}

sin (x), cos (x), 0, \frac{π}{2}

(x - 1) y^{^{'}} - 2 + 3 x + y = 0

What is (\partial)/(\partial x)(2^{-x}) ?
The answer to (\partial)/(\partial x)(2^{-x}) is -ln(2)*2^{-x}