What is the derivative of y= 6/(\sqrt[4]{x)} ?

The derivative of y= 6/(\sqrt[4]{x)} is -3/(2x^{5/4)}

What is the first derivative of y= 6/(\sqrt[4]{x)} ?

The first derivative of y= 6/(\sqrt[4]{x)} is -3/(2x^{5/4)}

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derivative of

- \frac{3}{2 x ^{\frac{5}{4}}}

Solution steps

Take the constant out:

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

Apply exponent rule:

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

= 6 \frac{d}{d x} (x^{- \frac{1}{4}})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 6 (- \frac{1}{4} x^{- \frac{1}{4} - 1})

Simplify

6 (- \frac{1}{4} x^{- \frac{1}{4} - 1}) : - \frac{3}{2 x ^{\frac{5}{4}}}

= - \frac{3}{2 x ^{\frac{5}{4}}}

t \to 0 lim (\frac{sin ( 3 t ) + 4 t}{t sec ( t )})

\frac{\partial}{\partial y} (5 x ln (x y))

\frac{\partial}{\partial x} (y^{2} cos (x))

\frac{\partial}{\partial x} (sin (9 x + yz))

\frac{d}{d x} (\frac{x ^{3} - 2 x + 4}{x ^{3}})

What is the derivative of y= 6/(\sqrt[4]{x)} ?
The derivative of y= 6/(\sqrt[4]{x)} is -3/(2x^{5/4)}
What is the first derivative of y= 6/(\sqrt[4]{x)} ?
The first derivative of y= 6/(\sqrt[4]{x)} is -3/(2x^{5/4)}