What is (\partial)/(\partial t)(t+sin(t-1)) ?

The answer to (\partial)/(\partial t)(t+sin(t-1)) is 1+cos(t-1)

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(\partial)/(\partial t)(t+sin(t-1))

\frac{\partial}{\partial t} (t + sin (t - 1))

1 + cos (t - 1)

Solution steps

\frac{\partial}{\partial t} (t + sin (t - 1))

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial t} (t) + \frac{\partial}{\partial t} (sin (t - 1))

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

\frac{\partial}{\partial t} (sin (t - 1)) = cos (t - 1)

= 1 + cos (t - 1)

\int 3 tan^{3} (2 x) sec^{5} (2 x) d x

x^{3} \cdot y^{^{'}} = x

[(z - 1)^{3} (z + 1)^{2}]

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

((\frac{1 + e ^{x}}{2})^{9})^{^{'}}

What is (\partial)/(\partial t)(t+sin(t-1)) ?
The answer to (\partial)/(\partial t)(t+sin(t-1)) is 1+cos(t-1)