What is the derivative of \sqrt[4]{x^2+y^2} ?

The derivative of \sqrt[4]{x^2+y^2} is (x+y\frac{dy)/(dx)}{2(x^2+y^2)^{3/4}}

What is the first derivative of \sqrt[4]{x^2+y^2} ?

The first derivative of \sqrt[4]{x^2+y^2} is (x+y\frac{dy)/(dx)}{2(x^2+y^2)^{3/4}}

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\frac{x + y \frac{d y}{d x}}{2 ( x ^{2} + y ^{2} ) ^{\frac{3}{4}}}

Solution steps

Apply the chain rule:

\frac{1}{4 ( x ^{2} + y ^{2} ) ^{\frac{3}{4}}} \frac{d}{d x} (x^{2} + y^{2})

= \frac{1}{4 ( x ^{2} + y ^{2} ) ^{\frac{3}{4}}} \frac{d}{d x} (x^{2} + y^{2})

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

= \frac{1}{4 ( x ^{2} + y ^{2} ) ^{\frac{3}{4}}} (2 x + 2 y \frac{d y}{d x})

Simplify

\frac{1}{4 ( x ^{2} + y ^{2} ) ^{\frac{3}{4}}} (2 x + 2 y \frac{d y}{d x}) : \frac{x + y \frac{d y}{d x}}{2 ( x ^{2} + y ^{2} ) ^{\frac{3}{4}}}

= \frac{x + y \frac{d y}{d x}}{2 ( x ^{2} + y ^{2} ) ^{\frac{3}{4}}}

\frac{d}{d x} (\frac{x ^{4} - x + 1}{x ^{4} + x + 1})

\int cos (x^{3} - 1) d x

x \to 2 lim (- x^{2} + 4 x + 2)

\int (- 2 sin (2 x) + 2 cos (2 x)) d x

h \to 0 lim (\frac{cos ( \frac{π}{6} + h ) - \frac{3}{2}}{h})

What is the derivative of \sqrt[4]{x^2+y^2} ?
The derivative of \sqrt[4]{x^2+y^2} is (x+y\frac{dy)/(dx)}{2(x^2+y^2)^{3/4}}
What is the first derivative of \sqrt[4]{x^2+y^2} ?
The first derivative of \sqrt[4]{x^2+y^2} is (x+y\frac{dy)/(dx)}{2(x^2+y^2)^{3/4}}