What is the derivative of (2x^{(-3x)}) ?

The derivative of (2x^{(-3x)}) is 8^{-x}x^{-3x}(-3ln(2x)-3)

What is the first derivative of (2x^{(-3x)}) ?

The first derivative of (2x^{(-3x)}) is 8^{-x}x^{-3x}(-3ln(2x)-3)

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

8^{- x} x^{- 3 x} (- 3 ln (2 x) - 3)

Solution steps

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

Apply exponent rule:

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

(2 x)^{- 3 x} = e^{(- 3 x) l n (2 x)}

= \frac{d}{d x} (e^{(- 3 x) l n (2 x)})

Apply the chain rule:

e^{(- 3 x) l n (2 x)} \frac{d}{d x} ((- 3 x) ln (2 x))

= e^{(- 3 x) l n (2 x)} \frac{d}{d x} ((- 3 x) ln (2 x))

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

= e^{(- 3 x) l n (2 x)} (- 3 ln (2 x) - 3)

Simplify

e^{(- 3 x) l n (2 x)} (- 3 ln (2 x) - 3) : 8^{- x} x^{- 3 x} (- 3 ln (2 x) - 3)

= 8^{- x} x^{- 3 x} (- 3 ln (2 x) - 3)

y = 7 - x^{2}

y = (1 - x) (x^{2} - 8)^{2}, (3, - 2)

\frac{d x}{d t} = - x - x^{2}

ln (3 x + 2)

n \to \infty lim (i = 1 \sum n \frac{i ^{3}}{n ^{4}} + \frac{1}{n})

What is the derivative of (2x^{(-3x)}) ?
The derivative of (2x^{(-3x)}) is 8^{-x}x^{-3x}(-3ln(2x)-3)
What is the first derivative of (2x^{(-3x)}) ?
The first derivative of (2x^{(-3x)}) is 8^{-x}x^{-3x}(-3ln(2x)-3)