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

The answer to (\partial)/(\partial y)(4x^2+4xy-3y^2+4y-1) is 4x+4-6y

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(\partial)/(\partial y)(4x^2+4xy-3y^2+4y-1)

\frac{\partial}{\partial y} (4 x^{2} + 4 x y - 3 y^{2} + 4 y - 1)

4 x + 4 - 6 y

Solution steps

\frac{\partial}{\partial y} (4 x^{2} + 4 x y - 3 y^{2} + 4 y - 1)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (4 x^{2}) + \frac{\partial}{\partial y} (4 x y) - \frac{\partial}{\partial y} (3 y^{2}) + \frac{\partial}{\partial y} (4 y) - \frac{\partial}{\partial y} (1)

\frac{\partial}{\partial y} (4 x^{2}) = 0

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

\frac{\partial}{\partial y} (3 y^{2}) = 6 y

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

\frac{\partial}{\partial y} (1) = 0

= 0 + 4 x - 6 y + 4 - 0

Simplify

= 4 x + 4 - 6 y

x \to 1 lim (x - 1)

\int sec^{2} (x) + sec (2 x) tan (2 x) d x

p (θ) = 7 θ

f (x) = ln (x + 4)

\int \frac{1}{2 sin ( x ) - 3 cos ( x ) - 5} d x

What is (\partial)/(\partial y)(4x^2+4xy-3y^2+4y-1) ?
The answer to (\partial)/(\partial y)(4x^2+4xy-3y^2+4y-1) is 4x+4-6y