What is (\partial)/(\partial t)(1-cos(t)) ?

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

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(\partial)/(\partial t)(1-cos(t))

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

sin (t)

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 0 - (- sin (t))

Simplify

= sin (t)

\frac{d}{d x} (\frac{8}{x ^{7}})

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

\int_{0}^{1} t^{3} (1 + t^{4})^{3} d t

\frac{d y}{d x} = - 10 x^{4} + 8 x^{3}

f (x) = 7 x - x, (4, 26)

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