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

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

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

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

2 (t - y)

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 + 2 (t - y)

Simplify

= 2 (t - y)

\int \frac{x ^{3} + 4 x ^{2} + x - 1}{x ^{3} + x ^{2}} d x

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\frac{d}{d x} (\frac{- e ^{- x} x - e ^{- x}}{x ^{2}})

\int_{- 7}^{7} (1 + 49 - x^{2}) d x

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