What is the derivative of 6xarcsin(x) ?

The derivative of 6xarcsin(x) is 6(arcsin(x)+x/(sqrt(1-x^2)))

What is the first derivative of 6xarcsin(x) ?

The first derivative of 6xarcsin(x) is 6(arcsin(x)+x/(sqrt(1-x^2)))

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derivative of

6 x arcsin (x)

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

Solution steps

\frac{d}{d x} (6 x arcsin (x))

Take the constant out:

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

= 6 \frac{d}{d x} (x arcsin (x))

Apply the Product Rule:

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

f = x, g = arcsin (x)

= 6 (\frac{d x}{d x} arcsin (x) + \frac{d}{d x} (arcsin (x)) x)

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

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

= 6 (1 \cdot arcsin (x) + \frac{1}{1 - x ^{2}} x)

Simplify

6 (1 \cdot arcsin (x) + \frac{1}{1 - x ^{2}} x) : 6 (arcsin (x) + \frac{x}{1 - x ^{2}})

= 6 (arcsin (x) + \frac{x}{1 - x ^{2}})

(t + e^{t}) (9 - t)

3 y^{^{'''}} + 12 y^{^{''}} + 12 y^{^{'}} = 0

\int_{0}^{1} (x^{2} + x) e^{x} d x

\int_{0}^{1} e^{- t^{2}} d t

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 8 y = 0

What is the derivative of 6xarcsin(x) ?
The derivative of 6xarcsin(x) is 6(arcsin(x)+x/(sqrt(1-x^2)))
What is the first derivative of 6xarcsin(x) ?
The first derivative of 6xarcsin(x) is 6(arcsin(x)+x/(sqrt(1-x^2)))