What is (\partial)/(\partial x)(4x^{0.5}y^{0.5}) ?

The answer to (\partial)/(\partial x)(4x^{0.5}y^{0.5}) is (2y^{1/2})/(x^{0.5)}

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(\partial)/(\partial x)(4x^{0.5}y^{0.5})

\frac{\partial}{\partial x} (4 x^{0.5} y^{0.5})

\frac{2 y ^{\frac{1}{2}}}{x ^{0.5}}

Solution steps

\frac{\partial}{\partial x} (4 x^{0.5} y^{0.5})

Treat

y

as a constant

Take the constant out:

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

= 4 y^{0.5} \frac{\partial}{\partial x} (x^{0.5})

Apply the Power Rule:

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

= 4 y^{0.5} \cdot 0.5 x^{0.5 - 1}

Simplify

4 y^{\frac{1}{2}} \cdot 0.5 x^{0.5 - 1} : \frac{2 y ^{\frac{1}{2}}}{x ^{0.5}}

= \frac{2 y ^{\frac{1}{2}}}{x ^{0.5}}

y = sin^{2} (x^{3})

\int \frac{3 x ^{2}}{x ^{3} + 1} d x

h \to 0 lim (\frac{( h - 3 ) ^{2} - 9}{h})

8 + x^{2}, 1

f (x) = 4 x^{5} ln (x)

What is (\partial)/(\partial x)(4x^{0.5}y^{0.5}) ?
The answer to (\partial)/(\partial x)(4x^{0.5}y^{0.5}) is (2y^{1/2})/(x^{0.5)}