What is the derivative of 6x^{1/3}+3x^{4/3} ?

The derivative of 6x^{1/3}+3x^{4/3} is (2+4x)/(x^{2/3)}

What is the first derivative of 6x^{1/3}+3x^{4/3} ?

The first derivative of 6x^{1/3}+3x^{4/3} is (2+4x)/(x^{2/3)}

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\frac{d}{d x} (6 x^{\frac{1}{3}} + 3 x^{\frac{4}{3}})

\frac{2 + 4 x}{x ^{\frac{2}{3}}}

Solution steps

\frac{d}{d x} (6 x^{\frac{1}{3}} + 3 x^{\frac{4}{3}})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (6 x^{\frac{1}{3}}) + \frac{d}{d x} (3 x^{\frac{4}{3}})

\frac{d}{d x} (6 x^{\frac{1}{3}}) = \frac{2}{x ^{\frac{2}{3}}}

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

= \frac{2}{x ^{\frac{2}{3}}} + 4 x^{\frac{1}{3}}

Simplify

\frac{2}{x ^{\frac{2}{3}}} + 4 x^{\frac{1}{3}} : \frac{2 + 4 x}{x ^{\frac{2}{3}}}

= \frac{2 + 4 x}{x ^{\frac{2}{3}}}

\int \frac{x + 1}{x} d x

(x^{2} - 2 x + 3)^{^{'}}

\int_{- 1}^{0} x^{3} d x

- \frac{32}{x ^{3}} + \frac{4}{x ^{2}}

(1 + y e^{x y}) d x + (4 + x e^{x y}) d y = 0

What is the derivative of 6x^{1/3}+3x^{4/3} ?
The derivative of 6x^{1/3}+3x^{4/3} is (2+4x)/(x^{2/3)}
What is the first derivative of 6x^{1/3}+3x^{4/3} ?
The first derivative of 6x^{1/3}+3x^{4/3} is (2+4x)/(x^{2/3)}