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Popular >

derivative of 6\sqrt[4]{x^3}

Solution

Solution

\frac{9}{2 x ^{\frac{1}{4}}}

Solution steps

Take the constant out:

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

Apply the chain rule:

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

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

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

= 6 \cdot \frac{1}{4 ( x ^{3} ) ^{\frac{3}{4}}} \cdot 3 x^{2}

Simplify

6 \cdot \frac{1}{4 ( x ^{3} ) ^{\frac{3}{4}}} \cdot 3 x^{2} : \frac{9}{2 x ^{\frac{1}{4}}}

= \frac{9}{2 x ^{\frac{1}{4}}}

Popular Examples

derivative of 3/(5x^2)

\frac{d}{d x} (\frac{3}{5 x ^{2}})

integral of sin(x)sin(2x)cos(3x)

\int sin (x) sin (2 x) cos (3 x) d x

(\partial)/(\partial x)((-xy)/(x^2+y^2))

\frac{\partial}{\partial x} (\frac{- x y}{x ^{2} + y ^{2}})

limit as x approaches 3+of (3-x)/(x^2-9)

x \to 3 + lim (\frac{3 - x}{x ^{2} - 9})

limit as t approaches 1 of (t^2-t)/(t-1)

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

Frequently Asked Questions (FAQ)

What is the derivative of 6\sqrt[4]{x^3} ?
The derivative of 6\sqrt[4]{x^3} is 9/(2x^{1/4)}
What is the first derivative of 6\sqrt[4]{x^3} ?
The first derivative of 6\sqrt[4]{x^3} is 9/(2x^{1/4)}