What is (\partial)/(\partial x)(ysqrt(x^2+y^2)) ?

The answer to (\partial)/(\partial x)(ysqrt(x^2+y^2)) is (yx)/(sqrt(x^2+y^2))

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(\partial)/(\partial x)(ysqrt(x^2+y^2))

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to \infty lim (3 x^{3} - 1000 x^{2})

\int \frac{sin ( ln ( x ^{6} ) )}{x} d x

\int \frac{1}{x ^{2} - 2 x - 3} d x

f (x) = - \frac{1}{4} x^{2}

f^{^{'}} (x) = \frac{x}{4}

What is (\partial)/(\partial x)(ysqrt(x^2+y^2)) ?
The answer to (\partial)/(\partial x)(ysqrt(x^2+y^2)) is (yx)/(sqrt(x^2+y^2))