What is the derivative of y^2+x ?

The derivative of y^2+x is 2y(dy)/(dx)+1

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

The first derivative of y^2+x is 2y(dy)/(dx)+1

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

2 y \frac{d y}{d x} + 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d x}{d x} = 1

= 2 y \frac{d y}{d x} + 1

\frac{d}{d x} (\frac{- e ^{x} x + 2 e ^{x}}{( 1 - x ) ^{2}})

sin (3 x), \frac{π}{3}

\frac{d y}{d x} = \frac{x}{y ^{2} 1 + x}

\int 2 e^{2 y} d y

\int_{- 12}^{0} \frac{1}{4 - x} d x

What is the derivative of y^2+x ?
The derivative of y^2+x is 2y(dy)/(dx)+1
What is the first derivative of y^2+x ?
The first derivative of y^2+x is 2y(dy)/(dx)+1