What is the derivative of 3x\sqrt[4]{x^4-2} ?

The derivative of 3x\sqrt[4]{x^4-2} is 3(\sqrt[4]{x^4-2}+(x^4)/((x^4-2)^{3/4)})

What is the first derivative of 3x\sqrt[4]{x^4-2} ?

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

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Solution steps

Take the constant out:

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

\frac{d x}{d x} = 1

Simplify

\frac{d}{d x} (4 x^{3} + 2 + \frac{1}{x})

\frac{d y}{d x} = 2 x e^{5 y}

\frac{\partial}{\partial x} (x sin (2 x))

\frac{d}{d x} (21 x^{3} + \frac{9}{x})

\frac{\partial}{\partial x} (x e^{\frac{x}{y}})

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