What is the d/(dt)(t^2-sec(t)+t) ?

The d/(dt)(t^2-sec(t)+t) is 2t-sec(t)tan(t)+1

What is the first d/(dt)(t^2-sec(t)+t) ?

The first d/(dt)(t^2-sec(t)+t) is 2t-sec(t)tan(t)+1

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\frac{d}{d t} (t^{2} - sec (t) + t)

2 t - sec (t) tan (t) + 1

Solution steps

\frac{d}{d t} (t^{2} - sec (t) + t)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (t^{2}) - \frac{d}{d t} (sec (t)) + \frac{d t}{d t}

\frac{d}{d t} (t^{2}) = 2 t

\frac{d}{d t} (sec (t)) = sec (t) tan (t)

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

= 2 t - sec (t) tan (t) + 1

\frac{3}{2 - x}

5 x^{2} + 3 x + 2

ln (3 x^{2} + 4 x + 5)

\frac{( 1 - cos ( x ) )}{( 1 + cos ( x ) )}

\frac{d}{d t} (10 - 2 t)

What is the d/(dt)(t^2-sec(t)+t) ?
The d/(dt)(t^2-sec(t)+t) is 2t-sec(t)tan(t)+1
What is the first d/(dt)(t^2-sec(t)+t) ?
The first d/(dt)(t^2-sec(t)+t) is 2t-sec(t)tan(t)+1