What is the derivative of x^2+4y^2-4xy+2 ?

The derivative of x^2+4y^2-4xy+2 is 2x+8y(dy)/(dx)-4(x(dy)/(dx)+y)

What is the first derivative of x^2+4y^2-4xy+2 ?

The first derivative of x^2+4y^2-4xy+2 is 2x+8y(dy)/(dx)-4(x(dy)/(dx)+y)

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

2 x + 8 y \frac{d y}{d x} - 4 (x \frac{d y}{d x} + y)

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

\frac{d}{d x} (4 x y) = 4 (y + x \frac{d y}{d x})

\frac{d}{d x} (2) = 0

= 2 x + 8 y \frac{d y}{d x} - 4 (y + x \frac{d y}{d x}) + 0

Simplify

= 2 x + 8 y \frac{d y}{d x} - 4 (x \frac{d y}{d x} + y)

f (x) = (2 x + 1)^{5} + 10 x (2 x + 1)^{4}

y^{^{'}} + \frac{y ^{2}}{x ^{2}} = 0

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

\frac{d y}{d x} + 4 y = - x

\int e^{2 x} (2) d x

What is the derivative of x^2+4y^2-4xy+2 ?
The derivative of x^2+4y^2-4xy+2 is 2x+8y(dy)/(dx)-4(x(dy)/(dx)+y)
What is the first derivative of x^2+4y^2-4xy+2 ?
The first derivative of x^2+4y^2-4xy+2 is 2x+8y(dy)/(dx)-4(x(dy)/(dx)+y)