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)

\int \frac{( x ^{3} - 2 x ^{2} - 4 )}{( x ^{3} - 2 x ^{2} )} d x

\int \frac{x ^{4}}{4} d x

\frac{d y}{d x} = y (x y^{5} - 1)

x y^{^{'}} + y = 0

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)