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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial x} (3) = 0

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

= 0 - \frac{2 x}{3 ( x ^{2} + y ^{2} ) ^{\frac{2}{3}}}

Simplify

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

\int 7 e^{- 0.4 x} d x

(2 y + 1) = \frac{d y}{d t}

f (x) = 4 x^{- 2}

\int \frac{3}{4} - x^{2} d x

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