What is (\partial)/(\partial x)(x^3+2y^2+8x+5y+5xy) ?

The answer to (\partial)/(\partial x)(x^3+2y^2+8x+5y+5xy) is 3x^2+5y+8

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(\partial)/(\partial x)(x^3+2y^2+8x+5y+5xy)

\frac{\partial}{\partial x} (x^{3} + 2 y^{2} + 8 x + 5 y + 5 x y)

3 x^{2} + 5 y + 8

Solution steps

\frac{\partial}{\partial x} (x^{3} + 2 y^{2} + 8 x + 5 y + 5 x y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{3}) + \frac{\partial}{\partial x} (2 y^{2}) + \frac{\partial}{\partial x} (8 x) + \frac{\partial}{\partial x} (5 y) + \frac{\partial}{\partial x} (5 x y)

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

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

\frac{\partial}{\partial x} (8 x) = 8

\frac{\partial}{\partial x} (5 y) = 0

\frac{\partial}{\partial x} (5 x y) = 5 y

= 3 x^{2} + 0 + 8 + 0 + 5 y

Simplify

= 3 x^{2} + 5 y + 8

y^{^{''}} + 9 y^{^{'}} = 40 e^{1 t}

\int 5 e^{t} d t

x \to 0 lim (co tan (x))

x \to - 1 lim (x^{3} - 2 x^{2} + 3 x - 4)

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

What is (\partial)/(\partial x)(x^3+2y^2+8x+5y+5xy) ?
The answer to (\partial)/(\partial x)(x^3+2y^2+8x+5y+5xy) is 3x^2+5y+8