What is (\partial)/(\partial x)((x^2-y^2)^{0.5}) ?

The answer to (\partial)/(\partial x)((x^2-y^2)^{0.5}) is x/((x^2-y^2)^{1/2)}

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(\partial)/(\partial x)((x^2-y^2)^{0.5})

\frac{\partial}{\partial x} ((x^{2} - y^{2})^{0.5})

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

Solution steps

\frac{\partial}{\partial x} ((x^{2} - y^{2})^{0.5})

Treat

y

as a constant

Apply the chain rule:

\frac{0.5}{( x ^{2} - y ^{2} ) ^{0.5}} \frac{\partial}{\partial x} (x^{2} - y^{2})

= \frac{0.5}{( x ^{2} - y ^{2} ) ^{0.5}} \frac{\partial}{\partial x} (x^{2} - y^{2})

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

= \frac{0.5}{( x ^{2} - y ^{2} ) ^{0.5}} \cdot 2 x

Simplify

\frac{0.5}{( 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}}}

\int x 2 x + 1 d x

\frac{\partial}{\partial y} (- \frac{sin ( 2 x y )}{x y})

x^{- \frac{3}{4}} + x^{\frac{3}{4}}

\frac{d}{d x} (2 ln (2 x + 5))

\frac{d}{d x} (\frac{x}{1 + 2 x})

What is (\partial)/(\partial x)((x^2-y^2)^{0.5}) ?
The answer to (\partial)/(\partial x)((x^2-y^2)^{0.5}) is x/((x^2-y^2)^{1/2)}