What is the derivative of (-5/(\sqrt[4]{x^3)}) ?

The derivative of (-5/(\sqrt[4]{x^3)}) is (15)/(4x^{7/4)}

What is the first derivative of (-5/(\sqrt[4]{x^3)}) ?

The first derivative of (-5/(\sqrt[4]{x^3)}) is (15)/(4x^{7/4)}

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

\frac{15}{4 x ^{\frac{7}{4}}}

Solution steps

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

Simplify

\frac{- 5}{4 x ^{3}} : - \frac{5}{4 x ^{3}}

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

Take the constant out:

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

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

Apply exponent rule:

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

= - 5 \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}

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

Simplify

- 5 (- \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \cdot 3 x^{2}) : \frac{15}{4 x ^{\frac{7}{4}}}

= \frac{15}{4 x ^{\frac{7}{4}}}

in v erse l a pl a ce \frac{5 s + 5}{s ( s ^{2} + 6 s + 10 )}

x \to - 1 lim (- \frac{2}{( x + 1 ) ^{2}})

\frac{d}{d y} (x - y)

in v erse l a pl a ce \frac{2}{s + 1} + \frac{24 s}{s ^{2} + 9}

f^{^{''}} (x) = 4, f^{^{'}} (2) = 10, f (2) = 15

What is the derivative of (-5/(\sqrt[4]{x^3)}) ?
The derivative of (-5/(\sqrt[4]{x^3)}) is (15)/(4x^{7/4)}
What is the first derivative of (-5/(\sqrt[4]{x^3)}) ?
The first derivative of (-5/(\sqrt[4]{x^3)}) is (15)/(4x^{7/4)}