What is (\partial)/(\partial t)((3y^2-t^2)/(y^5)) ?

The answer to (\partial)/(\partial t)((3y^2-t^2)/(y^5)) is -(2t)/(y^5)

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(\partial)/(\partial t)((3y^2-t^2)/(y^5))

\frac{\partial}{\partial t} (\frac{3 y ^{2} - t ^{2}}{y ^{5}})

- \frac{2 t}{y ^{5}}

Solution steps

\frac{\partial}{\partial t} (\frac{3 y ^{2} - t ^{2}}{y ^{5}})

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{y ^{5}} \frac{\partial}{\partial t} (3 y^{2} - t^{2})

Apply the Sum/Difference Rule:

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

= \frac{1}{y ^{5}} (\frac{\partial}{\partial t} (3 y^{2}) - \frac{\partial}{\partial t} (t^{2}))

\frac{\partial}{\partial t} (3 y^{2}) = 0

\frac{\partial}{\partial t} (t^{2}) = 2 t

= \frac{1}{y ^{5}} (0 - 2 t)

Simplify

\frac{1}{y ^{5}} (0 - 2 t) : - \frac{2 t}{y ^{5}}

= - \frac{2 t}{y ^{5}}

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What is (\partial)/(\partial t)((3y^2-t^2)/(y^5)) ?
The answer to (\partial)/(\partial t)((3y^2-t^2)/(y^5)) is -(2t)/(y^5)