What is the derivative of f(x)=(12x^3-9x)^{4/3} ?

The derivative of f(x)=(12x^3-9x)^{4/3} is 12(4x^2-1)(12x^3-9x)^{1/3}

What is the first derivative of f(x)=(12x^3-9x)^{4/3} ?

The first derivative of f(x)=(12x^3-9x)^{4/3} is 12(4x^2-1)(12x^3-9x)^{1/3}

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derivative of

f (x) = (12 x^{3} - 9 x)^{\frac{4}{3}}

12 (4 x^{2} - 1) (12 x^{3} - 9 x)^{\frac{1}{3}}

Solution steps

\frac{d}{d x} ((12 x^{3} - 9 x)^{\frac{4}{3}})

Apply the chain rule:

\frac{4 ( 12 x ^{3} - 9 x ) ^{\frac{1}{3}}}{3} \frac{d}{d x} (12 x^{3} - 9 x)

= \frac{4 ( 12 x ^{3} - 9 x ) ^{\frac{1}{3}}}{3} \frac{d}{d x} (12 x^{3} - 9 x)

\frac{d}{d x} (12 x^{3} - 9 x) = 36 x^{2} - 9

= \frac{4 ( 12 x ^{3} - 9 x ) ^{\frac{1}{3}}}{3} (36 x^{2} - 9)

Simplify

\frac{4 ( 12 x ^{3} - 9 x ) ^{\frac{1}{3}}}{3} (36 x^{2} - 9) : 12 (4 x^{2} - 1) (12 x^{3} - 9 x)^{\frac{1}{3}}

= 12 (4 x^{2} - 1) (12 x^{3} - 9 x)^{\frac{1}{3}}

\frac{d}{d x} (x^{3} - 9 x)

\int - \frac{2 x}{1 + x ^{2}} d x

f (x) = \frac{4}{x}, at x = - 2

\int_{1}^{8} 5 \frac{ln ( x )}{x ^{2}} d x

x \to \infty lim (0. 9^{x})

What is the derivative of f(x)=(12x^3-9x)^{4/3} ?
The derivative of f(x)=(12x^3-9x)^{4/3} is 12(4x^2-1)(12x^3-9x)^{1/3}
What is the first derivative of f(x)=(12x^3-9x)^{4/3} ?
The first derivative of f(x)=(12x^3-9x)^{4/3} is 12(4x^2-1)(12x^3-9x)^{1/3}