What is (\partial)/(\partial x)(2xy^6+x^4y^8) ?

The answer to (\partial)/(\partial x)(2xy^6+x^4y^8) is 2y^6+4y^8x^3

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(\partial)/(\partial x)(2xy^6+x^4y^8)

\frac{\partial}{\partial x} (2 x y^{6} + x^{4} y^{8})

2 y^{6} + 4 y^{8} x^{3}

Solution steps

\frac{\partial}{\partial x} (2 x y^{6} + x^{4} y^{8})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (2 x y^{6}) + \frac{\partial}{\partial x} (x^{4} y^{8})

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

\frac{\partial}{\partial x} (x^{4} y^{8}) = 4 y^{8} x^{3}

= 2 y^{6} + 4 y^{8} x^{3}

f (x) = x^{2} - \frac{1}{x}, at x = 1

\int 8 sin^{5} (x) cos (x) d x

(2 - 3 x) cos (3 x)

f (t) = (2 t - 1)^{3}

What is (\partial)/(\partial x)(2xy^6+x^4y^8) ?
The answer to (\partial)/(\partial x)(2xy^6+x^4y^8) is 2y^6+4y^8x^3