What is the inverse of y= x/((sqrt(1-x^2))) ?

The inverse of y= x/((sqrt(1-x^2))) is (xsqrt(x^2+1))/(x^2+1),-(xsqrt(x^2+1))/(x^2+1)

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inverse of y= x/((sqrt(1-x^2)))

inverse

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

\frac{x x ^{2} + 1}{x ^{2} + 1}, - \frac{x x ^{2} + 1}{x ^{2} + 1}

Solution steps

y = \frac{x}{1 - x ^{2}}

Replace

x

with

y

x = \frac{y}{1 - y ^{2}}

Solve for

y, x = \frac{y}{1 - y ^{2}}

\frac{x x ^{2} + 1}{x ^{2} + 1}, - \frac{x x ^{2} + 1}{x ^{2} + 1}

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

f (x) = - \frac{1}{3 ( \frac{1}{3 x} )}

sin (0.925)

f (x) = 14 x + 2

f (x) = \frac{( 2 x + 7 )}{10 x - 12}

What is the inverse of y= x/((sqrt(1-x^2))) ?
The inverse of y= x/((sqrt(1-x^2))) is (xsqrt(x^2+1))/(x^2+1),-(xsqrt(x^2+1))/(x^2+1)