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

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

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

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

- 2 y t

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

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

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

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

= y (0 - 2 t + 0)

Simplify

= - 2 y t

\frac{v ^{2} + 13}{( v - 9 ) ( v + 4 )}

\frac{\partial}{\partial x} ((y) (2 x + y))

\frac{sin ( ( 2 k + 1 ) 2 π )}{2 k + 1}

\frac{- 2 s ^{3} + 6 s ^{2} - 17 s + 54}{( s ^{2} + 9 ) ^{2}}

\frac{2 ( 5 + 2 \cdot 5 ^{\frac{1}{2}} )}{5}

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