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

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

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

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

- a^{- x} ln (a)

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

Simplify

e^{(- x) l n (a)} (- ln (a)) : - a^{- x} ln (a)

= - a^{- x} ln (a)

\int_{0}^{π} 8 sin (\frac{x}{2}) d x

y^{^{''}} - 6 y^{^{'}} + 8 y = - 2 cos (4 t)

\int_{0}^{2} 8 t d t

f (t) = t

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