What is (\partial)/(\partial y)(x^{3/4}y^{1/4}) ?

The answer to (\partial)/(\partial y)(x^{3/4}y^{1/4}) is (x^{3/4})/(4y^{3/4)}

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(\partial)/(\partial y)(x^{3/4}y^{1/4})

\frac{\partial}{\partial y} (x^{\frac{3}{4}} y^{\frac{1}{4}})

\frac{x ^{\frac{3}{4}}}{4 y ^{\frac{3}{4}}}

Solution steps

\frac{\partial}{\partial y} (x^{\frac{3}{4}} y^{\frac{1}{4}})

Treat

x

as a constant

Take the constant out:

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

= x^{\frac{3}{4}} \frac{\partial}{\partial y} (y^{\frac{1}{4}})

Apply the Power Rule:

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

= x^{\frac{3}{4}} \frac{1}{4} y^{\frac{1}{4} - 1}

Simplify

x^{\frac{3}{4}} \frac{1}{4} y^{\frac{1}{4} - 1} : \frac{x ^{\frac{3}{4}}}{4 y ^{\frac{3}{4}}}

= \frac{x ^{\frac{3}{4}}}{4 y ^{\frac{3}{4}}}

y = x^{2}, y = 4

f (t) = e^{5 t s i n (2 t)}

x \to 2 lim (3 x^{4} + 4 x^{3} - x + 2)

4 y^{^{''}} - 4 y^{^{'}} + 13 y = 0

\frac{d y}{d x}, y = e^{4 x^{2} + 1}

What is (\partial)/(\partial y)(x^{3/4}y^{1/4}) ?
The answer to (\partial)/(\partial y)(x^{3/4}y^{1/4}) is (x^{3/4})/(4y^{3/4)}