What is the first (1-cos(t))^' ?

The first (1-cos(t))^' is sin(t)

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(1-cos(t))^'

(1 - cos (t))^{^{'}}

sin (t)

Solution steps

(1 - cos (t))^{'}

Apply the Sum/Difference Rule:

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

= 1^{^{'}} - (cos (t))^{^{'}}

1^{'} = 0

(cos (t))^{'} = - sin (t)

= 0 - (- sin (t))

Simplify

= sin (t)

\int_{0}^{2} 1 + 16 x^{2} d x

\int \frac{1}{x ^{15}} d x

x \to π lim (5 \frac{x - π}{x + π})

\frac{d y}{d x} + 4 y = 60

2 π \frac{x}{g}