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

The derivative of (\sqrt[3]{x}/(t-3)) is 1/(3x^{2/3)(t-3)}

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

The first derivative of (\sqrt[3]{x}/(t-3)) is 1/(3x^{2/3)(t-3)}

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\frac{d}{d x} (\frac{3 x}{t - 3})

\frac{1}{3 x ^{\frac{2}{3}} ( t - 3 )}

Solution steps

\frac{d}{d x} (\frac{3 x}{t - 3})

Treat

t

as a constant

Take the constant out:

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

= \frac{1}{t - 3} \frac{d}{d x} (3 x)

Apply radical rule:

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

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

Apply the Power Rule:

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

= \frac{1}{t - 3} \cdot \frac{1}{3} x^{\frac{1}{3} - 1}

Simplify

\frac{1}{t - 3} \cdot \frac{1}{3} x^{\frac{1}{3} - 1} : \frac{1}{3 x ^{\frac{2}{3}} ( t - 3 )}

= \frac{1}{3 x ^{\frac{2}{3}} ( t - 3 )}

\frac{\partial}{\partial x} (\frac{8 x y ^{2}}{x ^{2} + y ^{2}})

\frac{\partial}{\partial x} (2 \cdot (y + 4) \cdot (x - 4)^{3})

x \to - 3 lim (1.2 x + 2.6)

\frac{\partial}{\partial x} (3 x^{4} y^{5} + 6 x^{7} y^{8})

y = \frac{x - 1}{x - 2} + 3

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