What is the derivative of y=sqrt(1-x^2)arcsin(x) ?

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

What is the first derivative of y=sqrt(1-x^2)arcsin(x) ?

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

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derivative of

y = 1 - x^{2} arcsin (x)

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

Solution steps

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

Apply the Product Rule:

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

f = 1 - x^{2}, g = arcsin (x)

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

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

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

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

Simplify

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

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

y^{^{'}} = (x^{2})

y^{^{'}} = \frac{( 5 x sec ( \frac{y}{x} ) + y )}{x}

\frac{d}{d x} (\frac{y}{1 - 2 x})

\int \frac{e ^{x} + 1}{2} d x

\int_{a}^{b} \frac{1}{r} d r

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