What is the derivative of 1-\sqrt[3]{x^2+y^2} ?

The derivative of 1-\sqrt[3]{x^2+y^2} is -(2x+2y\frac{dy)/(dx)}{3(x^2+y^2)^{2/3}}

What is the first derivative of 1-\sqrt[3]{x^2+y^2} ?

The first derivative of 1-\sqrt[3]{x^2+y^2} is -(2x+2y\frac{dy)/(dx)}{3(x^2+y^2)^{2/3}}

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (1) = 0

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

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

Simplify

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

\int sec^{3} (π) x d x

\frac{d}{d x} ((x - 1)^{\frac{1}{3}})

y^{^{'}} = 3 y + 6

x \to 0 lim (\frac{x}{x - 2})

\frac{\partial}{\partial x} (x^{2} + y^{10})

What is the derivative of 1-\sqrt[3]{x^2+y^2} ?
The derivative of 1-\sqrt[3]{x^2+y^2} is -(2x+2y\frac{dy)/(dx)}{3(x^2+y^2)^{2/3}}
What is the first derivative of 1-\sqrt[3]{x^2+y^2} ?
The first derivative of 1-\sqrt[3]{x^2+y^2} is -(2x+2y\frac{dy)/(dx)}{3(x^2+y^2)^{2/3}}