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

The answer to (\partial)/(\partial x)(x-3y^2+4x^2y^3) is 8xy^3+1

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

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

8 x y^{3} + 1

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (x) = 1

\frac{\partial}{\partial x} (3 y^{2}) = 0

\frac{\partial}{\partial x} (4 x^{2} y^{3}) = 8 y^{3} x

= 1 - 0 + 8 y^{3} x

Simplify

= 8 x y^{3} + 1

4 x^{2} + 5

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

\frac{d}{d x} (e^{x} + ln (x))

y^{^{''}} - 4 y^{^{'}} = 6 t

\frac{7 s ^{3} - 9 s + 1}{3 s ^{5}}

What is (\partial)/(\partial x)(x-3y^2+4x^2y^3) ?
The answer to (\partial)/(\partial x)(x-3y^2+4x^2y^3) is 8xy^3+1