What is (\partial)/(\partial y)(40x^{4/5}y^{1/5}) ?

The answer to (\partial)/(\partial y)(40x^{4/5}y^{1/5}) is (8x^{4/5})/(y^{4/5)}

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(\partial)/(\partial y)(40x^{4/5}y^{1/5})

\frac{\partial}{\partial y} (40 x^{\frac{4}{5}} y^{\frac{1}{5}})

\frac{8 x ^{\frac{4}{5}}}{y ^{\frac{4}{5}}}

Solution steps

\frac{\partial}{\partial y} (40 x^{\frac{4}{5}} y^{\frac{1}{5}})

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 40 x^{\frac{4}{5}} \frac{\partial}{\partial y} (y^{\frac{1}{5}})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 40 x^{\frac{4}{5}} \frac{1}{5} y^{\frac{1}{5} - 1}

Simplify

40 x^{\frac{4}{5}} \frac{1}{5} y^{\frac{1}{5} - 1} : \frac{8 x ^{\frac{4}{5}}}{y ^{\frac{4}{5}}}

= \frac{8 x ^{\frac{4}{5}}}{y ^{\frac{4}{5}}}

\frac{d y}{d t} + y \cdot cot (t) = cos (t)

\frac{d}{d x} ((3 x^{2} - 6 x + 6) e^{x})

y = 2 x^{3} + x^{2} + 23

cos^{2} (2 θ)

y^{^{'}} = - 10 y, y (0) = 5

What is (\partial)/(\partial y)(40x^{4/5}y^{1/5}) ?
The answer to (\partial)/(\partial y)(40x^{4/5}y^{1/5}) is (8x^{4/5})/(y^{4/5)}