What is the derivative of y= 2/3 x^3-3/(4x^2) ?

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

What is the first derivative of y= 2/3 x^3-3/(4x^2) ?

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

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derivative of

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

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

Solution steps

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

Simplify

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

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

= 2 x^{2} + \frac{3}{2 x ^{3}}

\frac{\partial}{\partial y} (\frac{cos ( y )}{x z ^{2}})

y^{^{'''}} = 6

\int sin (3 π t) d t

\int \frac{1}{5 - 2 x} d x

6 x - 4 x^{3} + 9

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