What is (\partial)/(\partial x)((x^2+y^2)^{1/2}) ?

The answer to (\partial)/(\partial x)((x^2+y^2)^{1/2}) is x/((x^2+y^2)^{1/2)}

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{( 1 - sec ( x ) )}{tan ( x )})

x + \frac{1}{x ^{2}} + x^{3}

\int 5 e^{- 2 x} + 4 d x

6 \frac{d y}{d x} - y^{2} = 0

2 cos (t)

What is (\partial)/(\partial x)((x^2+y^2)^{1/2}) ?
The answer to (\partial)/(\partial x)((x^2+y^2)^{1/2}) is x/((x^2+y^2)^{1/2)}