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

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

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(t-cos(t))^'

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

1 + sin (t)

Solution steps

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

Apply the Sum/Difference Rule:

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

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

t^{'} = 1

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

= 1 - (- sin (t))

Simplify

= 1 + sin (t)

x \to 0 lim (\frac{1}{1 + sin ( x )})

\frac{d h}{d t} = - h

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

\frac{d}{d x} ((x + 1) (x + 2))

x \to 0 lim (x - 8)