What is the derivative of 16\sqrt[4]{4x^4+4} ?

The derivative of 16\sqrt[4]{4x^4+4} is (64x^3)/((4x^4+4)^{3/4)}

What is the first derivative of 16\sqrt[4]{4x^4+4} ?

The first derivative of 16\sqrt[4]{4x^4+4} is (64x^3)/((4x^4+4)^{3/4)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

16 4 4 x^{4} + 4

\frac{64 x ^{3}}{( 4 x ^{4} + 4 ) ^{\frac{3}{4}}}

Solution steps

\frac{d}{d x} (16 4 4 x^{4} + 4)

Take the constant out:

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

= 16 \frac{d}{d x} (4 4 x^{4} + 4)

Apply the chain rule:

\frac{1}{4 ( 4 x ^{4} + 4 ) ^{\frac{3}{4}}} \frac{d}{d x} (4 x^{4} + 4)

= \frac{1}{4 ( 4 x ^{4} + 4 ) ^{\frac{3}{4}}} \frac{d}{d x} (4 x^{4} + 4)

\frac{d}{d x} (4 x^{4} + 4) = 16 x^{3}

= 16 \cdot \frac{1}{4 ( 4 x ^{4} + 4 ) ^{\frac{3}{4}}} \cdot 16 x^{3}

Simplify

16 \cdot \frac{1}{4 ( 4 x ^{4} + 4 ) ^{\frac{3}{4}}} \cdot 16 x^{3} : \frac{64 x ^{3}}{( 4 x ^{4} + 4 ) ^{\frac{3}{4}}}

= \frac{64 x ^{3}}{( 4 x ^{4} + 4 ) ^{\frac{3}{4}}}

t^{3} - \frac{1}{4 t ^{7}}

\frac{\partial}{\partial y} (y^{2} sin (x))

f (x) = \frac{1}{x - 1}, at x = 2

\frac{1}{s ( s + 8 )}

\int x^{x} (1 + ln (x)) d x

What is the derivative of 16\sqrt[4]{4x^4+4} ?
The derivative of 16\sqrt[4]{4x^4+4} is (64x^3)/((4x^4+4)^{3/4)}
What is the first derivative of 16\sqrt[4]{4x^4+4} ?
The first derivative of 16\sqrt[4]{4x^4+4} is (64x^3)/((4x^4+4)^{3/4)}