What is the d/(da)(2sin(2a)+2cos^2(a)) ?

The d/(da)(2sin(2a)+2cos^2(a)) is 4cos(2a)-2sin(2a)

What is the first d/(da)(2sin(2a)+2cos^2(a)) ?

The first d/(da)(2sin(2a)+2cos^2(a)) is 4cos(2a)-2sin(2a)

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\frac{d}{d a} (2 sin (2 a) + 2 cos^{2} (a))

4 cos (2 a) - 2 sin (2 a)

Solution steps

\frac{d}{d a} (2 sin (2 a) + 2 cos^{2} (a))

Apply the Sum/Difference Rule:

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

= \frac{d}{d a} (2 sin (2 a)) + \frac{d}{d a} (2 cos^{2} (a))

\frac{d}{d a} (2 sin (2 a)) = 4 cos (2 a)

\frac{d}{d a} (2 cos^{2} (a)) = - 2 sin (2 a)

= 4 cos (2 a) - 2 sin (2 a)

\frac{2}{s + 1} + \frac{16 s}{s ^{2} + 9}

(1.4) (3.5)

\frac{1}{s ^{2} + 3 s}

\frac{d}{d x} ((2 x)^{8})

x \to 0 + lim (6 - ln (x))

What is the d/(da)(2sin(2a)+2cos^2(a)) ?
The d/(da)(2sin(2a)+2cos^2(a)) is 4cos(2a)-2sin(2a)
What is the first d/(da)(2sin(2a)+2cos^2(a)) ?
The first d/(da)(2sin(2a)+2cos^2(a)) is 4cos(2a)-2sin(2a)