What is (\partial)/(\partial x)(x^2-4xy) ?

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

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(\partial)/(\partial x)(x^2-4xy)

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

2 x - 4 y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 2 x - 4 y

\int (- 0.98 wl x - 0.98 w l^{2}) d x

3 sin (t) cos (t)

n = 1 \sum \infty (0.2)^{n}

\int_{\frac{π}{2}}^{π} cos^{2} (x) d x

2 y^{^{''}} + 18 y = 0

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