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

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

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

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

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

1 + cos (t)

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d t}{d t} = 1

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

= 1 + cos (t)

\int \frac{1}{5} e^{5 x} d x

n = 1 \sum \infty 4 (\frac{3}{2})^{n - 1}

\frac{d y}{d x} = \frac{y ( x - y )}{x ^{2}}

\frac{d}{d x} (sin (x) \cdot cos (x) \cdot \frac{1}{x})

y = 6 x^{2}, y = 4 x^{2}, y = 6 - 5 x