What is the derivative of 4sin^2(0.5t)+1 ?

The derivative of 4sin^2(0.5t)+1 is 4sin(0.5t)cos(0.5t)

What is the first derivative of 4sin^2(0.5t)+1 ?

The first derivative of 4sin^2(0.5t)+1 is 4sin(0.5t)cos(0.5t)

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derivative of

4 sin^{2} (0.5 t) + 1

4 sin (0.5 t) cos (0.5 t)

Solution steps

\frac{d}{d t} (4 sin^{2} (0.5 t) + 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (4 sin^{2} (0.5 t)) + \frac{d}{d t} (1)

\frac{d}{d t} (4 sin^{2} (0.5 t)) = 4 sin (0.5 t) cos (0.5 t)

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

= 4 sin (0.5 t) cos (0.5 t) + 0

Simplify

= 4 sin (0.5 t) cos (0.5 t)

f (x) = 3 x, at x = 4

\int \frac{4 x ^{2} - 49}{x ^{3}} d x

\frac{x}{5 x + 2}

\int \frac{6 e ^{6 t}}{3 + e ^{6 t}} d t

x \to \infty lim (e^{x} + x)

What is the derivative of 4sin^2(0.5t)+1 ?
The derivative of 4sin^2(0.5t)+1 is 4sin(0.5t)cos(0.5t)
What is the first derivative of 4sin^2(0.5t)+1 ?
The first derivative of 4sin^2(0.5t)+1 is 4sin(0.5t)cos(0.5t)