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

The answer to (\partial)/(\partial x)(3x^2+2y^2-4xy+5) is 6x-4y

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(\partial)/(\partial x)(3x^2+2y^2-4xy+5)

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

6 x - 4 y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

\frac{\partial}{\partial x} (5) = 0

= 6 x + 0 - 4 y + 0

Simplify

= 6 x - 4 y

\frac{d}{d t} (\frac{t}{1 + ( 1.5574 + 5.43 t ^{2} ) ^{2}})

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

y^{^{'}} = 0.0014 y (1600 - y)

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

\int 3 - 4 x d x

What is (\partial)/(\partial x)(3x^2+2y^2-4xy+5) ?
The answer to (\partial)/(\partial x)(3x^2+2y^2-4xy+5) is 6x-4y