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

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

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

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

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

f (x) = 4 x^{\frac{4}{3}} - \frac{2}{3} x^{\frac{3}{2}} + x^{2} - 4 x + 1

n = 0 \sum \infty 2 \cdot 0. 2^{n}

x \to 0 lim (\frac{5 ^{x + 1} - 5}{x})

\int_{0}^{\infty} 8 sin^{2} (x) d x

\int arctan (d) x d x

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