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

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

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

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

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

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

Solution steps

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

Apply the Product Rule:

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

f = x^{\frac{1}{3}}, g = x + 4

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

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

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

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

Simplify

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

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

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

(24 + x^{2}) y^{^{'}} - 2 x y = 0

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

\int \frac{x ^{2} - 5 x + 9}{x ^{2} - 5 x + 6} d x

\frac{\partial}{\partial x} (x, y, z)

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