What is the derivative of y^2x ?

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

What is the first derivative of y^2x ?

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

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

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

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = y^{2}, g = x

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

\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} x + 1 \cdot y^{2}

Simplify

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

\int x (1 - x^{2}) d x

\int (x + sin (x))^{2} d x

t sin^{2} (t)

\int sec^{2} (θ) tan^{4} (θ) d θ

2 y^{^{'}} = y^{2}, y (0) = 7

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