What is the derivative of 2xy+y^2 ?

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

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

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

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

2 (y + x \frac{d y}{d x}) + 2 y \frac{d y}{d x}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

y = x - x^{2}, (- 1, - 2)

t^{3} sin (3 t)

f^{^{'}} (x) = 5 x + 1

(tan (y))^{^{'}}

x^{^{'}} = - 3 \cdot \frac{x}{300}

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