What is the (1/2 arcsin(x)+1/4 sin(2arcsin(x)))^' ?

The (1/2 arcsin(x)+1/4 sin(2arcsin(x)))^' is (1+cos(2arcsin(x)))/(2sqrt(-x^2+1))

What is the first (1/2 arcsin(x)+1/4 sin(2arcsin(x)))^' ?

The first (1/2 arcsin(x)+1/4 sin(2arcsin(x)))^' is (1+cos(2arcsin(x)))/(2sqrt(-x^2+1))

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(\frac{1}{2} arcsin (x) + \frac{1}{4} sin (2 arcsin (x)))^{^{'}}

\frac{1 + cos ( 2 arcsin ( x ) )}{2 - x ^{2} + 1}

Solution steps

(\frac{1}{2} arcsin (x) + \frac{1}{4} sin (2 arcsin (x)))^{'}

Apply the Sum/Difference Rule:

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

= (\frac{1}{2} arcsin (x))^{^{'}} + (\frac{1}{4} sin (2 arcsin (x)))^{^{'}}

(\frac{1}{2} arcsin (x))^{'} = \frac{1}{2 1 - x ^{2}}

(\frac{1}{4} sin (2 arcsin (x)))^{'} = \frac{cos ( 2 arcsin ( x ) )}{2 - x ^{2} + 1}

= \frac{1}{2 1 - x ^{2}} + \frac{cos ( 2 arcsin ( x ) )}{2 - x ^{2} + 1}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 + cos ( 2 arcsin ( x ) )}{2 - x ^{2} + 1}

x \cdot g (x) \cdot f (x)

f (x) = (6 x^{2} - x + 7)^{- \frac{5}{3}}

\frac{d y}{d t} = k (y - 60)

\int \frac{x}{( 4 x + 5 ) ^{2}} d x

x \to 1 lim ((x^{3} + 2) (3 - x))

What is the (1/2 arcsin(x)+1/4 sin(2arcsin(x)))^' ?
The (1/2 arcsin(x)+1/4 sin(2arcsin(x)))^' is (1+cos(2arcsin(x)))/(2sqrt(-x^2+1))
What is the first (1/2 arcsin(x)+1/4 sin(2arcsin(x)))^' ?
The first (1/2 arcsin(x)+1/4 sin(2arcsin(x)))^' is (1+cos(2arcsin(x)))/(2sqrt(-x^2+1))