What is (\partial)/(\partial x)(x^3*y^2+4x^2-3xy) ?

The answer to (\partial)/(\partial x)(x^3*y^2+4x^2-3xy) is 3y^2x^2+8x-3y

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(\partial)/(\partial x)(x^3*y^2+4x^2-3xy)

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

3 y^{2} x^{2} + 8 x - 3 y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (4 x^{2}) = 8 x

\frac{\partial}{\partial x} (3 x y) = 3 y

= 3 y^{2} x^{2} + 8 x - 3 y

x \to \infty lim (x - {x})

\int 2 yz d y

x \to 0 lim (\frac{( 2 + x ) ^{2} - 8}{x})

y = x + \frac{4}{x}, (- 4, - 5)

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

What is (\partial)/(\partial x)(x^3*y^2+4x^2-3xy) ?
The answer to (\partial)/(\partial x)(x^3*y^2+4x^2-3xy) is 3y^2x^2+8x-3y