What is the d/(dθ)(1/4 sin^2(2θ)) ?

The d/(dθ)(1/4 sin^2(2θ)) is (sin(4θ))/2

What is the first d/(dθ)(1/4 sin^2(2θ)) ?

The first d/(dθ)(1/4 sin^2(2θ)) is (sin(4θ))/2

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\frac{d}{d θ} (\frac{1}{4} sin^{2} (2 θ))

\frac{sin ( 4 θ )}{2}

Solution steps

\frac{d}{d θ} (\frac{1}{4} sin^{2} (2 θ))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{4} \frac{d}{d θ} (sin^{2} (2 θ))

Apply the chain rule:

2 sin (2 θ) \frac{d}{d θ} (sin (2 θ))

= 2 sin (2 θ) \frac{d}{d θ} (sin (2 θ))

\frac{d}{d θ} (sin (2 θ)) = cos (2 θ) \cdot 2

= \frac{1}{4} \cdot 2 sin (2 θ) cos (2 θ) \cdot 2

Simplify

\frac{1}{4} \cdot 2 sin (2 θ) cos (2 θ) \cdot 2 : \frac{sin ( 4 θ )}{2}

= \frac{sin ( 4 θ )}{2}

What is the d/(dθ)(1/4 sin^2(2θ)) ?
The d/(dθ)(1/4 sin^2(2θ)) is (sin(4θ))/2
What is the first d/(dθ)(1/4 sin^2(2θ)) ?
The first d/(dθ)(1/4 sin^2(2θ)) is (sin(4θ))/2