What is the first d/(dt)(1+sin(t)) ?

The first d/(dt)(1+sin(t)) is cos(t)

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d/(dt)(1+sin(t))

\frac{d}{d t} (1 + sin (t))

cos (t)

Solution steps

\frac{d}{d t} (1 + sin (t))

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (1) + \frac{d}{d t} (sin (t))

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

\frac{d}{d t} (sin (t)) = cos (t)

= 0 + cos (t)

Simplify

= cos (t)

\int - \frac{3}{x} d x

(x + 3)^{2}

\frac{d}{d y} (\frac{x}{( x ^{2} + y ^{2} )})

n = 1 \sum \infty \frac{( - 1 ) ^{n - 1} n ^{2}}{1 2 ^{n}}