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

The derivative of f(x)=(2x)^{4x} is ln(2)*2^{4x+2}x^{4x}+2^{2+4x}x^{4x}(ln(x)+1)

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

The first derivative of f(x)=(2x)^{4x} is ln(2)*2^{4x+2}x^{4x}+2^{2+4x}x^{4x}(ln(x)+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

f (x) = (2 x)^{4 x}

ln (2) \cdot 2^{4 x + 2} x^{4 x} + 2^{2 + 4 x} x^{4 x} (ln (x) + 1)

Solution steps

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

Simplify

(2 x)^{4 x} : 1 6^{x} x^{4 x}

= \frac{d}{d x} (1 6^{x} x^{4 x})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 1 6^{x}, g = x^{4 x}

= \frac{d}{d x} (1 6^{x}) x^{4 x} + \frac{d}{d x} (x^{4 x}) 1 6^{x}

\frac{d}{d x} (1 6^{x}) = ln (2) \cdot 2^{4 x + 2}

\frac{d}{d x} (x^{4 x}) = 4 x^{4 x} (ln (x) + 1)

= ln (2) \cdot 2^{4 x + 2} x^{4 x} + 4 x^{4 x} (ln (x) + 1) \cdot 1 6^{x}

4 x^{4 x} (ln (x) + 1) \cdot 1 6^{x} = 2^{2 + 4 x} x^{4 x} (ln (x) + 1)

= ln (2) \cdot 2^{4 x + 2} x^{4 x} + 2^{2 + 4 x} x^{4 x} (ln (x) + 1)

y^{^{''}} + 16 y = 2 sin (2 t)

\frac{d}{d x} (x^{2} - 5)

\int - 13 cos^{3} (x) sin (x) d x

\int_{0}^{1} \frac{5}{x ( x + 1 )} d x

(2 x y + 2 x^{3} y) d x - (1 + x^{2}) d y = 0

What is the derivative of f(x)=(2x)^{4x} ?
The derivative of f(x)=(2x)^{4x} is ln(2)*2^{4x+2}x^{4x}+2^{2+4x}x^{4x}(ln(x)+1)
What is the first derivative of f(x)=(2x)^{4x} ?
The first derivative of f(x)=(2x)^{4x} is ln(2)*2^{4x+2}x^{4x}+2^{2+4x}x^{4x}(ln(x)+1)