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

The answer to (\partial)/(\partial x)((x-5)^{1/4}y^{1/4}) is (y^{1/4})/(4(x-5)^{3/4)}

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

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

\frac{y ^{\frac{1}{4}}}{4 ( x - 5 ) ^{\frac{3}{4}}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

\frac{1}{4 ( x - 5 ) ^{\frac{3}{4}}} \frac{\partial}{\partial x} (x - 5)

= \frac{1}{4 ( x - 5 ) ^{\frac{3}{4}}} \frac{\partial}{\partial x} (x - 5)

\frac{\partial}{\partial x} (x - 5) = 1

= y^{\frac{1}{4}} \frac{1}{4 ( x - 5 ) ^{\frac{3}{4}}} \cdot 1

Simplify

y^{\frac{1}{4}} \frac{1}{4 ( x - 5 ) ^{\frac{3}{4}}} \cdot 1 : \frac{y ^{\frac{1}{4}}}{4 ( x - 5 ) ^{\frac{3}{4}}}

= \frac{y ^{\frac{1}{4}}}{4 ( x - 5 ) ^{\frac{3}{4}}}

f (x) = e^{x} sin (x)

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

\frac{d}{d x} (a^{2} - x^{2} + a arcsin (\frac{x}{a}))

\int x - \frac{1}{2} x d x

What is (\partial)/(\partial x)((x-5)^{1/4}y^{1/4}) ?
The answer to (\partial)/(\partial x)((x-5)^{1/4}y^{1/4}) is (y^{1/4})/(4(x-5)^{3/4)}