What is (\partial)/(\partial x)(10^{-2x}) ?

The answer to (\partial)/(\partial x)(10^{-2x}) is -ln(10)*25^{-x}*2^{-2x+1}

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(\partial)/(\partial x)(10^{-2x})

\frac{\partial}{\partial x} (1 0^{- 2 x})

- ln (10) \cdot 2 5^{- x} \cdot 2^{- 2 x + 1}

Solution steps

\frac{\partial}{\partial x} (1 0^{- 2 x})

Simplify

1 0^{- 2 x} : 10 0^{- x}

= \frac{\partial}{\partial x} (10 0^{- x})

Apply exponent rule:

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

10 0^{- x} = e^{(- x) l n (100)}

= \frac{\partial}{\partial x} (e^{(- x) l n (100)})

Apply the chain rule:

e^{(- x) l n (100)} \frac{\partial}{\partial x} ((- x) ln (100))

= e^{(- x) l n (100)} \frac{\partial}{\partial x} ((- x) ln (100))

\frac{\partial}{\partial x} ((- x) ln (100)) = - 2 ln (10)

= e^{(- x) l n (100)} (- 2 ln (10))

Simplify

e^{(- x) l n (100)} (- 2 ln (10)) : - ln (10) \cdot 2 5^{- x} \cdot 2^{- 2 x + 1}

= - ln (10) \cdot 2 5^{- x} \cdot 2^{- 2 x + 1}

\int a rcco tan (x) d x

\frac{d}{d x} (6 x^{3} x csc (x))

a re a x^{2}, x + 2, - 1, 2

\int sin^{5} (2 t) cos (2 t) d t

a re a 3 x^{3} - 2 x, - 3, 3

What is (\partial)/(\partial x)(10^{-2x}) ?
The answer to (\partial)/(\partial x)(10^{-2x}) is -ln(10)*25^{-x}*2^{-2x+1}