What is the derivative of 7/(4\sqrt[4]{x^3}) ?

The derivative of 7/(4\sqrt[4]{x^3}) is -(21)/(16x^{7/4)}

What is the first derivative of 7/(4\sqrt[4]{x^3}) ?

The first derivative of 7/(4\sqrt[4]{x^3}) is -(21)/(16x^{7/4)}

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\frac{d}{d x} (\frac{7}{4 4 x ^{3}})

- \frac{21}{16 x ^{\frac{7}{4}}}

Solution steps

\frac{d}{d x} (\frac{7}{4 4 x ^{3}})

Take the constant out:

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

= \frac{7}{4} \frac{d}{d x} (\frac{1}{4 x ^{3}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{7}{4} \frac{d}{d x} ((x^{3})^{- \frac{1}{4}})

Apply the chain rule:

- \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \frac{d}{d x} (x^{3})

= - \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \frac{d}{d x} (x^{3})

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

= \frac{7}{4} (- \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \cdot 3 x^{2})

Simplify

\frac{7}{4} (- \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \cdot 3 x^{2}) : - \frac{21}{16 x ^{\frac{7}{4}}}

= - \frac{21}{16 x ^{\frac{7}{4}}}

f (x) = \frac{( 13 x + 100 )}{x}

- x^{2} + 3, 3 x^{2} - 9 x

\int_{0}^{a} x^{2} a^{2} - x^{2} d x

\int_{1}^{e} 4 x ln (x) d x

(\frac{d y}{d x}) + 8 y = 12

What is the derivative of 7/(4\sqrt[4]{x^3}) ?
The derivative of 7/(4\sqrt[4]{x^3}) is -(21)/(16x^{7/4)}
What is the first derivative of 7/(4\sqrt[4]{x^3}) ?
The first derivative of 7/(4\sqrt[4]{x^3}) is -(21)/(16x^{7/4)}