What is (\partial)/(\partial x)(6xy-2+10x^2y^2) ?

The answer to (\partial)/(\partial x)(6xy-2+10x^2y^2) is 20xy^2+6y

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(\partial)/(\partial x)(6xy-2+10x^2y^2)

\frac{\partial}{\partial x} (6 x y - 2 + 10 x^{2} y^{2})

20 x y^{2} + 6 y

Solution steps

\frac{\partial}{\partial x} (6 x y - 2 + 10 x^{2} y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (6 x y) - \frac{\partial}{\partial x} (2) + \frac{\partial}{\partial x} (10 x^{2} y^{2})

\frac{\partial}{\partial x} (6 x y) = 6 y

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

\frac{\partial}{\partial x} (10 x^{2} y^{2}) = 20 y^{2} x

= 6 y - 0 + 20 y^{2} x

Simplify

= 20 x y^{2} + 6 y

\frac{\partial}{\partial x} (x arctan (y + z))

f (x) = \frac{x}{- 6 x + 6}, at x = 2

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

x \to 5 lim (\frac{1}{( x - 5 ) ^{2}})

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

What is (\partial)/(\partial x)(6xy-2+10x^2y^2) ?
The answer to (\partial)/(\partial x)(6xy-2+10x^2y^2) is 20xy^2+6y