What is the derivative of (x^2+y^2^{4/5}) ?

The derivative of (x^2+y^2^{4/5}) is (4(2x+2y\frac{dy)/(dx))}{5(x^2+y^2)^{1/5}}

What is the first derivative of (x^2+y^2^{4/5}) ?

The first derivative of (x^2+y^2^{4/5}) is (4(2x+2y\frac{dy)/(dx))}{5(x^2+y^2)^{1/5}}

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\frac{d}{d x} ((x^{2} + y^{2})^{\frac{4}{5}})

\frac{4 ( 2 x + 2 y \frac{d y}{d x} )}{5 ( x ^{2} + y ^{2} ) ^{\frac{1}{5}}}

Solution steps

\frac{d}{d x} ((x^{2} + y^{2})^{\frac{4}{5}})

Apply the chain rule:

\frac{4}{5 ( x ^{2} + y ^{2} ) ^{\frac{1}{5}}} \frac{d}{d x} (x^{2} + y^{2})

= \frac{4}{5 ( x ^{2} + y ^{2} ) ^{\frac{1}{5}}} \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{4}{5 ( x ^{2} + y ^{2} ) ^{\frac{1}{5}}} (2 x + 2 y \frac{d y}{d x})

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{4 ( 2 x + 2 y \frac{d y}{d x} )}{5 ( x ^{2} + y ^{2} ) ^{\frac{1}{5}}}

\int e^{3 x} - 1 d x

x \to \infty lim (e^{x} + e)

\frac{d}{d x} (3 x + x)

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

\frac{d}{d x} (a e^{x} cos (x) + b e^{x} sin (x))

What is the derivative of (x^2+y^2^{4/5}) ?
The derivative of (x^2+y^2^{4/5}) is (4(2x+2y\frac{dy)/(dx))}{5(x^2+y^2)^{1/5}}
What is the first derivative of (x^2+y^2^{4/5}) ?
The first derivative of (x^2+y^2^{4/5}) is (4(2x+2y\frac{dy)/(dx))}{5(x^2+y^2)^{1/5}}