What is the inverse of 1/((y^2+1)^{1/2)}y ?

The inverse of 1/((y^2+1)^{1/2)}y is sqrt(-(y^2)/(y^2-1)),-sqrt(-(y^2)/(y^2-1))

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inverse of 1/((y^2+1)^{1/2)}y

inverse

\frac{1}{( y ^{2} + 1 ) ^{\frac{1}{2}}} y

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

Solution steps

y = \frac{1}{( y ^{2} + 1 ) ^{\frac{1}{2}}} y

Replace

y

with

y

y = \frac{1}{( y ^{2} + 1 ) ^{\frac{1}{2}}} y

Solve for

y, y = \frac{1}{( y ^{2} + 1 ) ^{\frac{1}{2}}} y

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

g (x) = - 1 + x^{3}

f (x) = lo g_{2} (2 x - 10)

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

f (x) = 8 (\frac{x ^{\frac{1}{3}}}{9} + 4)

f (x) = x^{2} + 2 x - 8

What is the inverse of 1/((y^2+1)^{1/2)}y ?
The inverse of 1/((y^2+1)^{1/2)}y is sqrt(-(y^2)/(y^2-1)),-sqrt(-(y^2)/(y^2-1))