What is (\partial)/(\partial x)(3x^3y-4/x+2y^{1/2}) ?

The answer to (\partial)/(\partial x)(3x^3y-4/x+2y^{1/2}) is 9x^2y+4/(x^2)

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(\partial)/(\partial x)(3x^3y-4/x+2y^{1/2})

\frac{\partial}{\partial x} (3 x^{3} y - \frac{4}{x} + 2 y^{\frac{1}{2}})

9 x^{2} y + \frac{4}{x ^{2}}

Solution steps

\frac{\partial}{\partial x} (3 x^{3} y - \frac{4}{x} + 2 y^{\frac{1}{2}})

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (3 x^{3} y) - \frac{\partial}{\partial x} (\frac{4}{x}) + \frac{\partial}{\partial x} (2 y^{\frac{1}{2}})

\frac{\partial}{\partial x} (3 x^{3} y) = 9 y x^{2}

\frac{\partial}{\partial x} (\frac{4}{x}) = - \frac{4}{x ^{2}}

\frac{\partial}{\partial x} (2 y^{\frac{1}{2}}) = 0

= 9 y x^{2} - (- \frac{4}{x ^{2}}) + 0

Simplify

= 9 x^{2} y + \frac{4}{x ^{2}}

\int_{\frac{1}{7}}^{3} 8 x ln (7 x) d x

y = 3 x + 6 cos (x), at (\frac{π}{3})

\frac{s}{s ^{2} + 2 s + 3}

f (x) = \frac{2 x ^{3} - 3 x + 1}{x}

y = x^{2} - 2

What is (\partial)/(\partial x)(3x^3y-4/x+2y^{1/2}) ?
The answer to (\partial)/(\partial x)(3x^3y-4/x+2y^{1/2}) is 9x^2y+4/(x^2)