What is (\partial)/(\partial u)((u^2+v^2)^{3/2}) ?

The answer to (\partial)/(\partial u)((u^2+v^2)^{3/2}) is 3u(u^2+v^2)^{1/2}

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(\partial)/(\partial u)((u^2+v^2)^{3/2})

\frac{\partial}{\partial u} ((u^{2} + v^{2})^{\frac{3}{2}})

3 u (u^{2} + v^{2})^{\frac{1}{2}}

Solution steps

\frac{\partial}{\partial u} ((u^{2} + v^{2})^{\frac{3}{2}})

Treat

v

as a constant

Apply the chain rule:

\frac{3 ( u ^{2} + v ^{2} ) ^{\frac{1}{2}}}{2} \frac{\partial}{\partial u} (u^{2} + v^{2})

= \frac{3 ( u ^{2} + v ^{2} ) ^{\frac{1}{2}}}{2} \frac{\partial}{\partial u} (u^{2} + v^{2})

\frac{\partial}{\partial u} (u^{2} + v^{2}) = 2 u

= \frac{3 ( u ^{2} + v ^{2} ) ^{\frac{1}{2}}}{2} \cdot 2 u

Simplify

\frac{3 ( u ^{2} + v ^{2} ) ^{\frac{1}{2}}}{2} \cdot 2 u : 3 u (u^{2} + v^{2})^{\frac{1}{2}}

= 3 u (u^{2} + v^{2})^{\frac{1}{2}}

x \to 4 - lim (\frac{1}{x - 4})

\int (x^{2} - 1)^{- 1} d x

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

x^{2} + x - 2, [- 3, 4]

\int_{1}^{2} 3 x^{4} ln (x) d x

What is (\partial)/(\partial u)((u^2+v^2)^{3/2}) ?
The answer to (\partial)/(\partial u)((u^2+v^2)^{3/2}) is 3u(u^2+v^2)^{1/2}