What is (\partial)/(\partial x)(98y^2+x^2-x^2y) ?

The answer to (\partial)/(\partial x)(98y^2+x^2-x^2y) is 2x-2yx

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(\partial)/(\partial x)(98y^2+x^2-x^2y)

\frac{\partial}{\partial x} (98 y^{2} + x^{2} - x^{2} y)

2 x - 2 y x

Solution steps

\frac{\partial}{\partial x} (98 y^{2} + x^{2} - x^{2} y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (98 y^{2}) + \frac{\partial}{\partial x} (x^{2}) - \frac{\partial}{\partial x} (x^{2} y)

\frac{\partial}{\partial x} (98 y^{2}) = 0

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

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

= 0 + 2 x - 2 y x

Simplify

= 2 x - 2 y x

\frac{d}{d x} ((x - 4)^{2} - 5)

n = 0 \sum \infty n e^{- n^{2}}

f (x) = (1 + 3 x^{2}) (x - x^{2})

y^{^{'}} + 2 y = t e^{- 2 t}, y (1) = 0

\int [(3 - x^{2})^{3}] d x

What is (\partial)/(\partial x)(98y^2+x^2-x^2y) ?
The answer to (\partial)/(\partial x)(98y^2+x^2-x^2y) is 2x-2yx