What is the derivative of f(x)=(3x^4+8)^{4/3} ?

The derivative of f(x)=(3x^4+8)^{4/3} is 16x^3(3x^4+8)^{1/3}

What is the first derivative of f(x)=(3x^4+8)^{4/3} ?

The first derivative of f(x)=(3x^4+8)^{4/3} is 16x^3(3x^4+8)^{1/3}

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derivative of

f (x) = (3 x^{4} + 8)^{\frac{4}{3}}

16 x^{3} (3 x^{4} + 8)^{\frac{1}{3}}

Solution steps

\frac{d}{d x} ((3 x^{4} + 8)^{\frac{4}{3}})

Apply the chain rule:

\frac{4 ( 3 x ^{4} + 8 ) ^{\frac{1}{3}}}{3} \frac{d}{d x} (3 x^{4} + 8)

= \frac{4 ( 3 x ^{4} + 8 ) ^{\frac{1}{3}}}{3} \frac{d}{d x} (3 x^{4} + 8)

\frac{d}{d x} (3 x^{4} + 8) = 12 x^{3}

= \frac{4 ( 3 x ^{4} + 8 ) ^{\frac{1}{3}}}{3} \cdot 12 x^{3}

Simplify

\frac{4 ( 3 x ^{4} + 8 ) ^{\frac{1}{3}}}{3} \cdot 12 x^{3} : 16 x^{3} (3 x^{4} + 8)^{\frac{1}{3}}

= 16 x^{3} (3 x^{4} + 8)^{\frac{1}{3}}

\int_{0}^{1} 2 (1 + x)^{4} d x

\frac{d}{d x} (cos (x^{2}) - 2 x^{2} sin (x^{2}))

\frac{d}{d x} (1 - tan^{2} (x))

y = 1 - cos^{2} (x)

\frac{d}{d x} (- sin (\frac{x}{2}))

What is the derivative of f(x)=(3x^4+8)^{4/3} ?
The derivative of f(x)=(3x^4+8)^{4/3} is 16x^3(3x^4+8)^{1/3}
What is the first derivative of f(x)=(3x^4+8)^{4/3} ?
The first derivative of f(x)=(3x^4+8)^{4/3} is 16x^3(3x^4+8)^{1/3}