What is the derivative of (8x^2-40/(3x^{1/3)}) ?

The derivative of (8x^2-40/(3x^{1/3)}) is (40x^2+40)/(9x^{4/3)}

What is the first derivative of (8x^2-40/(3x^{1/3)}) ?

The first derivative of (8x^2-40/(3x^{1/3)}) is (40x^2+40)/(9x^{4/3)}

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\frac{d}{d x} (\frac{8 x ^{2} - 40}{3 x ^{\frac{1}{3}}})

\frac{40 x ^{2} + 40}{9 x ^{\frac{4}{3}}}

Solution steps

\frac{d}{d x} (\frac{8 x ^{2} - 40}{3 x ^{\frac{1}{3}}})

Take the constant out:

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

= \frac{1}{3} \frac{d}{d x} (\frac{8 x ^{2} - 40}{x ^{\frac{1}{3}}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{3} \cdot \frac{\frac{d}{d x} ( 8 x ^{2} - 40 ) x ^{\frac{1}{3}} - \frac{d}{d x} ( x ^{\frac{1}{3}} ) ( 8 x ^{2} - 40 )}{( x ^{\frac{1}{3}} ) ^{2}}

\frac{d}{d x} (8 x^{2} - 40) = 16 x

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

= \frac{1}{3} \cdot \frac{16 x x ^{\frac{1}{3}} - \frac{1}{3 x ^{\frac{2}{3}}} ( 8 x ^{2} - 40 )}{( x ^{\frac{1}{3}} ) ^{2}}

Simplify

\frac{1}{3} \cdot \frac{16 x x ^{\frac{1}{3}} - \frac{1}{3 x ^{\frac{2}{3}}} ( 8 x ^{2} - 40 )}{( x ^{\frac{1}{3}} ) ^{2}} : \frac{40 x ^{2} + 40}{9 x ^{\frac{4}{3}}}

= \frac{40 x ^{2} + 40}{9 x ^{\frac{4}{3}}}

x \to 0 + lim (\frac{∣ 2 x ∣}{x})

n \to \infty lim (A r g (- 1 + \frac{i}{n}))

x \to \infty + lim (\frac{11}{2})

\frac{0.5}{s \cdot ( s ^{2} + s + 1 )}

x \to 0 lim ((1 + 6 x)^{\frac{5}{x}})

What is the derivative of (8x^2-40/(3x^{1/3)}) ?
The derivative of (8x^2-40/(3x^{1/3)}) is (40x^2+40)/(9x^{4/3)}
What is the first derivative of (8x^2-40/(3x^{1/3)}) ?
The first derivative of (8x^2-40/(3x^{1/3)}) is (40x^2+40)/(9x^{4/3)}