What is the derivative of 2/3 x+4/(3x^2) ?

The derivative of 2/3 x+4/(3x^2) is 2/3-8/(3x^3)

What is the first derivative of 2/3 x+4/(3x^2) ?

The first derivative of 2/3 x+4/(3x^2) is 2/3-8/(3x^3)

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

\frac{2}{3} - \frac{8}{3 x ^{3}}

Solution steps

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

Simplify

\frac{2}{3} x + \frac{4}{3 x ^{2}} : \frac{2 x}{3} + \frac{4}{3 x ^{2}}

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (\frac{4}{3 x ^{2}}) = - \frac{8}{3 x ^{3}}

= \frac{2}{3} - \frac{8}{3 x ^{3}}

\frac{d}{d x} (f (x) \circ g (x))

x \to 1 - lim (\frac{x ^{3} - 1}{x - 1})

(x - 7)^{4} y^{^{'}} + 5 (x - 7)^{3} y = x + 7

\int \frac{1}{tan ( y )} d y

y^{^{''}} - 8 y^{^{'}} + 16 y = t^{- 7} e^{4 t}

What is the derivative of 2/3 x+4/(3x^2) ?
The derivative of 2/3 x+4/(3x^2) is 2/3-8/(3x^3)
What is the first derivative of 2/3 x+4/(3x^2) ?
The first derivative of 2/3 x+4/(3x^2) is 2/3-8/(3x^3)