What is the derivative of y= 9/(8\sqrt[4]{x^3)} ?

The derivative of y= 9/(8\sqrt[4]{x^3)} is -(27x^2)/(32(x^3)^{5/4)}

What is the first derivative of y= 9/(8\sqrt[4]{x^3)} ?

The first derivative of y= 9/(8\sqrt[4]{x^3)} is -(27x^2)/(32(x^3)^{5/4)}

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derivative of

- \frac{27 x ^{2}}{32 ( x ^{3} ) ^{\frac{5}{4}}}

Solution steps

Take the constant out:

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

Apply exponent rule:

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

= \frac{9}{8} \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{9}{8} (- \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \cdot 3 x^{2})

Simplify

\frac{9}{8} (- \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \cdot 3 x^{2}) : - \frac{27 x ^{2}}{32 ( x ^{3} ) ^{\frac{5}{4}}}

= - \frac{27 x ^{2}}{32 ( x ^{3} ) ^{\frac{5}{4}}}

x \to 0 lim (a^{x})

x^{5} - 7 x + 8

y^{^{''}} - 4 y^{^{'}} + 3 y = cos (4 t)

y = \frac{x}{x + 3}

\frac{d}{d x} (cos (\frac{x}{4}))

What is the derivative of y= 9/(8\sqrt[4]{x^3)} ?
The derivative of y= 9/(8\sqrt[4]{x^3)} is -(27x^2)/(32(x^3)^{5/4)}
What is the first derivative of y= 9/(8\sqrt[4]{x^3)} ?
The first derivative of y= 9/(8\sqrt[4]{x^3)} is -(27x^2)/(32(x^3)^{5/4)}