Question

1) Find d/dt ( sec t/2)

Answer 1

Answer:

$\frac{d}{dt} [sec(\frac{t}{2} )] = \frac{1}{2} sec(\frac{t}{2} )tan(\frac{t}{2} )$

General Formulas and Concepts:

Calculus

• Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$
• Trig u derivative: $\frac{d}{dt} [sec(u)] = u'[sec(u)tan(u)]$

Step-by-step explanation:

Step 1: Define

$\frac{d}{dt} [sec(\frac{t}{2} )]$

Step 2: Differentiate

1. Trig u [Chain Rule/Basic Power]:                                     $\frac{d}{dt} [sec(\frac{t}{2} )] = \frac{t^{1-1}}{2} sec(\frac{t}{2} )tan(\frac{t}{2} )$
2. Simplify:                                                                                 $\frac{d}{dt} [sec(\frac{t}{2} )] = \frac{t^{0}}{2} sec(\frac{t}{2} )tan(\frac{t}{2} )$
3. Evaluate:                                                                                    $\frac{d}{dt} [sec(\frac{t}{2} )] = \frac{1}{2} sec(\frac{t}{2} )tan(\frac{t}{2} )$