What is the derivative of (4x^{-2x}) ?

The derivative of (4x^{-2x}) is 16^{-x}x^{-2x}(-2ln(4x)-2)

What is the first derivative of (4x^{-2x}) ?

The first derivative of (4x^{-2x}) is 16^{-x}x^{-2x}(-2ln(4x)-2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

1 6^{- x} x^{- 2 x} (- 2 ln (4 x) - 2)

Solution steps

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

Apply exponent rule:

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

(4 x)^{- 2 x} = e^{(- 2 x) l n (4 x)}

= \frac{d}{d x} (e^{(- 2 x) l n (4 x)})

Apply the chain rule:

e^{(- 2 x) l n (4 x)} \frac{d}{d x} ((- 2 x) ln (4 x))

= e^{(- 2 x) l n (4 x)} \frac{d}{d x} ((- 2 x) ln (4 x))

\frac{d}{d x} ((- 2 x) ln (4 x)) = - 2 ln (4 x) - 2

= e^{(- 2 x) l n (4 x)} (- 2 ln (4 x) - 2)

Simplify

e^{(- 2 x) l n (4 x)} (- 2 ln (4 x) - 2) : 1 6^{- x} x^{- 2 x} (- 2 ln (4 x) - 2)

= 1 6^{- x} x^{- 2 x} (- 2 ln (4 x) - 2)

y = sin (sin (x))

y^{^{''}} = - 4 y

\int_{0}^{1} \frac{3}{2 t - 5} d t

\int_{1}^{4} x^{2} - 4 x - 3 d x

\int_{- \infty}^{4} x e^{x} d x

What is the derivative of (4x^{-2x}) ?
The derivative of (4x^{-2x}) is 16^{-x}x^{-2x}(-2ln(4x)-2)
What is the first derivative of (4x^{-2x}) ?
The first derivative of (4x^{-2x}) is 16^{-x}x^{-2x}(-2ln(4x)-2)