What is (\partial)/(\partial t)(t^2+1) ?

The answer to (\partial)/(\partial t)(t^2+1) is 2t

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(\partial)/(\partial t)(t^2+1)

\frac{\partial}{\partial t} (t^{2} + 1)

2 t

Solution steps

\frac{\partial}{\partial t} (t^{2} + 1)

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial t} (t^{2}) + \frac{\partial}{\partial t} (1)

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

\frac{\partial}{\partial t} (1) = 0

= 2 t + 0

Simplify

= 2 t

f (x) = \frac{( x ^{2} )}{x - 5}

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

\frac{u}{u ^{2} + 4}

e^{2 (t - 3)} (3 t^{4} + 2 t^{3} + t)

\frac{\partial}{\partial y} (cos (x - y) - x e^{y})

What is (\partial)/(\partial t)(t^2+1) ?
The answer to (\partial)/(\partial t)(t^2+1) is 2t