What is (\partial)/(\partial x)(4x^2-8xy^4+7ay^5-3) ?

The answer to (\partial)/(\partial x)(4x^2-8xy^4+7ay^5-3) is 8x-8y^4

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(\partial)/(\partial x)(4x^2-8xy^4+7ay^5-3)

\frac{\partial}{\partial x} (4 x^{2} - 8 x y^{4} + 7 a y^{5} - 3)

8 x - 8 y^{4}

Solution steps

\frac{\partial}{\partial x} (4 x^{2} - 8 x y^{4} + 7 a y^{5} - 3)

Treat

a, y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (4 x^{2}) - \frac{\partial}{\partial x} (8 x y^{4}) + \frac{\partial}{\partial x} (7 a y^{5}) - \frac{\partial}{\partial x} (3)

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

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

\frac{\partial}{\partial x} (7 a y^{5}) = 0

\frac{\partial}{\partial x} (3) = 0

= 8 x - 8 y^{4} + 0 - 0

Simplify

= 8 x - 8 y^{4}

sin (x + 1)

\frac{d}{d x} (- \frac{1}{2} cot (x))

\frac{d y}{d x} = e^{x} y^{2}

y = \frac{2 x + 5}{3 x - 2}

\int \frac{e ^{7 x}}{e ^{7 x} + 1} d x

What is (\partial)/(\partial x)(4x^2-8xy^4+7ay^5-3) ?
The answer to (\partial)/(\partial x)(4x^2-8xy^4+7ay^5-3) is 8x-8y^4