What is (\partial)/(\partial x)(40x^{3/4}y^{1/4}) ?

The answer to (\partial)/(\partial x)(40x^{3/4}y^{1/4}) is (30y^{1/4})/(x^{1/4)}

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(\partial)/(\partial x)(40x^{3/4}y^{1/4})

\frac{\partial}{\partial x} (40 x^{\frac{3}{4}} y^{\frac{1}{4}})

\frac{30 y ^{\frac{1}{4}}}{x ^{\frac{1}{4}}}

Solution steps

\frac{\partial}{\partial x} (40 x^{\frac{3}{4}} y^{\frac{1}{4}})

Treat

y

as a constant

Take the constant out:

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

= 40 y^{\frac{1}{4}} \frac{\partial}{\partial x} (x^{\frac{3}{4}})

Apply the Power Rule:

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

= 40 y^{\frac{1}{4}} \frac{3}{4} x^{\frac{3}{4} - 1}

Simplify

40 y^{\frac{1}{4}} \frac{3}{4} x^{\frac{3}{4} - 1} : \frac{30 y ^{\frac{1}{4}}}{x ^{\frac{1}{4}}}

= \frac{30 y ^{\frac{1}{4}}}{x ^{\frac{1}{4}}}

y = x^{3}, x = - 1

f (x) = (x^{3} + 1) (x^{2} - 2), at x = 2

f (x) = (x - 4) (3 - x)^{3}, at x = 5

\int x^{2} (3 x + x^{- 3}) d x

\int sin (- 3 x) d x

What is (\partial)/(\partial x)(40x^{3/4}y^{1/4}) ?
The answer to (\partial)/(\partial x)(40x^{3/4}y^{1/4}) is (30y^{1/4})/(x^{1/4)}