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

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

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

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

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derivative of

f (x) = \frac{x ^{\frac{2}{3}}}{x ^{\frac{1}{3}} + 2}

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{tan ( x ) - 5}{sec ( x )})

\frac{d x}{d y} = \frac{x - 11}{3 y + 2513}, y (0) = 77

t \to 2 lim (3 - t + 7)

f (x) = \frac{sin ( x )}{1 - cos ( x )}

\frac{5}{1 + x ^{2}}

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