What is the derivative of 9xarcsin(x) ?

The derivative of 9xarcsin(x) is 9(arcsin(x)+x/(sqrt(1-x^2)))

What is the first derivative of 9xarcsin(x) ?

The first derivative of 9xarcsin(x) is 9(arcsin(x)+x/(sqrt(1-x^2)))

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\frac{d}{d x} (9 x arcsin (x))

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

Solution steps

\frac{d}{d x} (9 x arcsin (x))

Take the constant out:

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

= 9 \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)

= 9 (\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}}

= 9 (1 \cdot arcsin (x) + \frac{1}{1 - x ^{2}} x)

Simplify

9 (1 \cdot arcsin (x) + \frac{1}{1 - x ^{2}} x) : 9 (arcsin (x) + \frac{x}{1 - x ^{2}})

= 9 (arcsin (x) + \frac{x}{1 - x ^{2}})

\frac{d}{d x} (6 x - 5)

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

\frac{1}{( s + 4 ) ^{2}}

y^{^{''}} - 2 y^{^{'}} + y = t + 1, y (0) = 0, y^{^{'}} (0) = 0

\frac{d}{d x} (5^{x + 1})

What is the derivative of 9xarcsin(x) ?
The derivative of 9xarcsin(x) is 9(arcsin(x)+x/(sqrt(1-x^2)))
What is the first derivative of 9xarcsin(x) ?
The first derivative of 9xarcsin(x) is 9(arcsin(x)+x/(sqrt(1-x^2)))