What is the derivative of 6^{-2x} ?

The derivative of 6^{-2x} is -ln(6)*9^{-x}*2^{-2x+1}

What is the first derivative of 6^{-2x} ?

The first derivative of 6^{-2x} is -ln(6)*9^{-x}*2^{-2x+1}

\frac{d}{d x} (6^{- 2 x})

- ln (6) \cdot 9^{- x} \cdot 2^{- 2 x + 1}

Solution steps

\frac{d}{d x} (6^{- 2 x})

Simplify

6^{- 2 x} : 3 6^{- x}

= \frac{d}{d x} (3 6^{- x})

Apply exponent rule:

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

3 6^{- x} = e^{(- x) l n (36)}

= \frac{d}{d x} (e^{(- x) l n (36)})

Apply the chain rule:

e^{(- x) l n (36)} \frac{d}{d x} ((- x) ln (36))

= e^{(- x) l n (36)} \frac{d}{d x} ((- x) ln (36))

\frac{d}{d x} ((- x) ln (36)) = - 2 ln (6)

= e^{(- x) l n (36)} (- 2 ln (6))

Simplify

e^{(- x) l n (36)} (- 2 ln (6)) : - ln (6) \cdot 9^{- x} \cdot 2^{- 2 x + 1}

= - ln (6) \cdot 9^{- x} \cdot 2^{- 2 x + 1}

\int ((x^{2} + 2 x) cos (x)) d x

d er i v a t i v e f (x) = 4 lo g_{6} (2)

t an g e n t f (x) = \frac{1}{2 x}, at x = 65

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

d er i v a t i v e f (x) = \frac{e ^{x}}{x ^{6}}

What is the derivative of 6^{-2x} ?
The derivative of 6^{-2x} is -ln(6)*9^{-x}*2^{-2x+1}
What is the first derivative of 6^{-2x} ?
The first derivative of 6^{-2x} is -ln(6)*9^{-x}*2^{-2x+1}