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)

h \to 0 lim (\frac{cos ( 6 h ) - 1}{h})

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

y^{^{''}} + y = 2 + x^{2}

\frac{d ^{2} y}{d x ^{2}} = 49 y

\frac{d}{d x} (6^{4} x^{3} + x^{5} - 180 4 x^{2})

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