What is the ((m^2+2)/(sqrt(2m^3+4)))^' ?

The ((m^2+2)/(sqrt(2m^3+4)))^' is 0

What is the first ((m^2+2)/(sqrt(2m^3+4)))^' ?

The first ((m^2+2)/(sqrt(2m^3+4)))^' is 0

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(\frac{m ^{2} + 2}{2 m ^{3} + 4})^{^{'}}

0

Solution steps

(\frac{m ^{2} + 2}{2 m ^{3} + 4})^{'}

Derivative of a constant:

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

= 0

y^{^{''}} + 9 y = 4 cos (2 x)

\frac{\partial}{\partial x} (2 e^{y})

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

x \to 0 lim ((1)^{\frac{1}{x}})

y^{^{''}} - 4 y = 9 e^{t}, y (0) = 0, y^{^{'}} (0) = 0

What is the ((m^2+2)/(sqrt(2m^3+4)))^' ?
The ((m^2+2)/(sqrt(2m^3+4)))^' is 0
What is the first ((m^2+2)/(sqrt(2m^3+4)))^' ?
The first ((m^2+2)/(sqrt(2m^3+4)))^' is 0