What is (\partial)/(\partial y)(yarcsin(xy)) ?

The answer to (\partial)/(\partial y)(yarcsin(xy)) is arcsin(yx)+(yx)/(sqrt(1-y^2x^2))

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(\partial)/(\partial y)(yarcsin(xy))

\frac{\partial}{\partial y} (y arcsin (x y))

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

Solution steps

\frac{\partial}{\partial y} (y arcsin (x y))

Treat

x

as a constant

Apply the Product Rule:

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

f = y, g = arcsin (x y)

= \frac{\partial}{\partial y} (y) arcsin (x y) + \frac{\partial}{\partial y} (arcsin (x y)) y

\frac{\partial}{\partial y} (y) = 1

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

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

Simplify

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

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

What is (\partial)/(\partial y)(yarcsin(xy)) ?
The answer to (\partial)/(\partial y)(yarcsin(xy)) is arcsin(yx)+(yx)/(sqrt(1-y^2x^2))