What is (\partial)/(\partial y)((x^2+y^2)^{4/5}) ?

The answer to (\partial)/(\partial y)((x^2+y^2)^{4/5}) is (8y)/(5(x^2+y^2)^{1/5)}

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(\partial)/(\partial y)((x^2+y^2)^{4/5})

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

\frac{8 y}{5 ( x ^{2} + y ^{2} ) ^{\frac{1}{5}}}

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

\frac{4}{5 ( x ^{2} + y ^{2} ) ^{\frac{1}{5}}} \cdot 2 y : \frac{8 y}{5 ( x ^{2} + y ^{2} ) ^{\frac{1}{5}}}

= \frac{8 y}{5 ( x ^{2} + y ^{2} ) ^{\frac{1}{5}}}

f (x) = x^{2} - x + 13, at x = 4

f (x) = - \frac{4}{3} x^{\frac{2}{3}}

(y^{2} + 1) d x = y \cdot sec^{2} (x) d y

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

y^{^{'}} + \frac{1}{x} y = 8 x^{2}

What is (\partial)/(\partial y)((x^2+y^2)^{4/5}) ?
The answer to (\partial)/(\partial y)((x^2+y^2)^{4/5}) is (8y)/(5(x^2+y^2)^{1/5)}