What is the derivative of arcsin(xdx) ?

The derivative of arcsin(xdx) is d(x/(sqrt(1-x^2))+arcsin(x))

What is the first derivative of arcsin(xdx) ?

The first derivative of arcsin(xdx) is d(x/(sqrt(1-x^2))+arcsin(x))

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\frac{d}{d x} (arcsin (x) d x)

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

Solution steps

\frac{d}{d x} (arcsin (x) d x)

Treat

d

as a constant

Take the constant out:

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

= d \frac{d}{d x} (arcsin (x) x)

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = arcsin (x), g = x

= d (\frac{d}{d x} (arcsin (x)) x + \frac{d x}{d x} arcsin (x))

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

\frac{d x}{d x} = 1

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

Simplify

d (\frac{1}{1 - x ^{2}} x + 1 \cdot arcsin (x)) : d (\frac{x}{1 - x ^{2}} + arcsin (x))

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

f (x) = (x^{3} - 25 x)^{8}, (- 5, 0)

\int \frac{2 x}{1 - x} d x

\int \frac{- 3 x}{( x - 1 ) ^{2} + 4} d x

y^{^{'''}} - y^{^{''}} + y^{^{'}} + 3 y = 0

\frac{d y}{d t} = 3 y + y^{4}, y (0) = 1

What is the derivative of arcsin(xdx) ?
The derivative of arcsin(xdx) is d(x/(sqrt(1-x^2))+arcsin(x))
What is the first derivative of arcsin(xdx) ?
The first derivative of arcsin(xdx) is d(x/(sqrt(1-x^2))+arcsin(x))