What is the derivative of f(x)=x^{4/3}-4x^{1/3}+1 ?

The derivative of f(x)=x^{4/3}-4x^{1/3}+1 is (4x-4)/(3x^{2/3)}

What is the first derivative of f(x)=x^{4/3}-4x^{1/3}+1 ?

The first derivative of f(x)=x^{4/3}-4x^{1/3}+1 is (4x-4)/(3x^{2/3)}

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derivative of

f (x) = x^{\frac{4}{3}} - 4 x^{\frac{1}{3}} + 1

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

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

\frac{d}{d x} (1) = 0

= \frac{4 x ^{\frac{1}{3}}}{3} - \frac{4}{3 x ^{\frac{2}{3}}} + 0

Simplify

\frac{4 x ^{\frac{1}{3}}}{3} - \frac{4}{3 x ^{\frac{2}{3}}} + 0 : \frac{4 x - 4}{3 x ^{\frac{2}{3}}}

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

\int_{0}^{π} sin (θ) d θ

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

x^{4} sec (\frac{5}{x})

z \to 0 lim (\frac{1}{f ( z ) \frac{1}{z}})

x + 4

What is the derivative of f(x)=x^{4/3}-4x^{1/3}+1 ?
The derivative of f(x)=x^{4/3}-4x^{1/3}+1 is (4x-4)/(3x^{2/3)}
What is the first derivative of f(x)=x^{4/3}-4x^{1/3}+1 ?
The first derivative of f(x)=x^{4/3}-4x^{1/3}+1 is (4x-4)/(3x^{2/3)}