What is the derivative of 73.3\sqrt[4]{x^3} ?

The derivative of 73.3\sqrt[4]{x^3} is (54.975)/(x^{1/4)}

What is the first derivative of 73.3\sqrt[4]{x^3} ?

The first derivative of 73.3\sqrt[4]{x^3} is (54.975)/(x^{1/4)}

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\frac{d}{d x} (73.3 4 x^{3})

\frac{54.975}{x ^{\frac{1}{4}}}

Solution steps

\frac{d}{d x} (73.3 4 x^{3})

Take the constant out:

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

= 73.3 \frac{d}{d x} (4 x^{3})

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}

= 73.3 \cdot \frac{1}{4 ( x ^{3} ) ^{\frac{3}{4}}} \cdot 3 x^{2}

Simplify

73.3 \cdot \frac{1}{4 ( x ^{3} ) ^{\frac{3}{4}}} \cdot 3 x^{2} : \frac{54.975}{x ^{\frac{1}{4}}}

= \frac{54.975}{x ^{\frac{1}{4}}}

\frac{d}{d t} (t^{3} - 12 t^{2} + 45 t)

l a pl a ce t r an s f or m 3 t

\int sin (6 θ) cos (2 θ) d θ

t an g e n t f (x) = (\frac{2 x}{x ^{2} + 1}), at x = 1

y^{^{''}} = - 3 y

What is the derivative of 73.3\sqrt[4]{x^3} ?
The derivative of 73.3\sqrt[4]{x^3} is (54.975)/(x^{1/4)}
What is the first derivative of 73.3\sqrt[4]{x^3} ?
The first derivative of 73.3\sqrt[4]{x^3} is (54.975)/(x^{1/4)}