What is (\partial)/(\partial y)(xsqrt(1+y^2)) ?

The answer to (\partial)/(\partial y)(xsqrt(1+y^2)) is (xy)/(sqrt(1+y^2))

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(\partial)/(\partial y)(xsqrt(1+y^2))

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

\frac{x y}{1 + y ^{2}}

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= x \frac{1}{2 1 + y ^{2}} \cdot 2 y

Simplify

x \frac{1}{2 1 + y ^{2}} \cdot 2 y : \frac{x y}{1 + y ^{2}}

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

\frac{\partial}{\partial y} (sin (x) cos (7 y))

\frac{\partial}{\partial u} (\frac{u - v}{2})

\frac{\partial}{\partial x} (5 x + y)

\int \frac{7 x ^{2}}{1 - x ^{6}} d x

\frac{d}{d y} (y z)

What is (\partial)/(\partial y)(xsqrt(1+y^2)) ?
The answer to (\partial)/(\partial y)(xsqrt(1+y^2)) is (xy)/(sqrt(1+y^2))