What is (\partial)/(\partial x)((x^3+y^2)^{1/2}) ?

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

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(\partial)/(\partial x)((x^3+y^2)^{1/2})

\frac{\partial}{\partial x} ((x^{3} + y^{2})^{\frac{1}{2}})

\frac{3 x ^{2}}{2 ( x ^{3} + y ^{2} ) ^{\frac{1}{2}}}

Solution steps

\frac{\partial}{\partial x} ((x^{3} + y^{2})^{\frac{1}{2}})

Treat

y

as a constant

Apply the chain rule:

\frac{1}{2 ( x ^{3} + y ^{2} ) ^{\frac{1}{2}}} \frac{\partial}{\partial x} (x^{3} + y^{2})

= \frac{1}{2 ( x ^{3} + y ^{2} ) ^{\frac{1}{2}}} \frac{\partial}{\partial x} (x^{3} + y^{2})

\frac{\partial}{\partial x} (x^{3} + y^{2}) = 3 x^{2}

= \frac{1}{2 ( x ^{3} + y ^{2} ) ^{\frac{1}{2}}} \cdot 3 x^{2}

Simplify

\frac{1}{2 ( x ^{3} + y ^{2} ) ^{\frac{1}{2}}} \cdot 3 x^{2} : \frac{3 x ^{2}}{2 ( x ^{3} + y ^{2} ) ^{\frac{1}{2}}}

= \frac{3 x ^{2}}{2 ( x ^{3} + y ^{2} ) ^{\frac{1}{2}}}

y^{^{''}} + 2 y^{^{'}} + 2 y = 1

\frac{d}{d x} (\frac{7 - sec ( x )}{tan ( x )})

x \to - \infty lim (\frac{x}{9 - x ^{2}})

x^{2} y^{^{''}} + 2 x y^{^{'}} - 72 y = 0

\int_{0}^{2} \frac{x + 1}{x ( x ^{2} + 1 )} d x

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