What is the derivative of 1/5 (xy^2+4y^2) ?

The derivative of 1/5 (xy^2+4y^2) is 2/5 (xy^2+4y)(y^2+2xy(dy)/(dx)+4(dy)/(dx))

What is the first derivative of 1/5 (xy^2+4y^2) ?

The first derivative of 1/5 (xy^2+4y^2) is 2/5 (xy^2+4y)(y^2+2xy(dy)/(dx)+4(dy)/(dx))

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\frac{d}{d x} (\frac{1}{5} (x y^{2} + 4 y)^{2})

\frac{2}{5} (x y^{2} + 4 y) (y^{2} + 2 x y \frac{d y}{d x} + 4 \frac{d y}{d x})

Solution steps

\frac{d}{d x} (\frac{1}{5} (x y^{2} + 4 y)^{2})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{5} \frac{d}{d x} ((x y^{2} + 4 y)^{2})

Apply the chain rule:

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

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

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

= \frac{1}{5} \cdot 2 (x y^{2} + 4 y) (y^{2} + 2 x y \frac{d y}{d x} + 4 \frac{d y}{d x})

= \frac{1}{5} \cdot 2 (x y^{2} + 4 y) (y^{2} + 2 x y \frac{d y}{d x} + 4 \frac{d y}{d x})

Simplify

\frac{1}{5} \cdot 2 (x y^{2} + 4 y) (y^{2} + 2 x y \frac{d y}{d x} + 4 \frac{d y}{d x}) : \frac{2}{5} (x y^{2} + 4 y) (y^{2} + 2 x y \frac{d y}{d x} + 4 \frac{d y}{d x})

= \frac{2}{5} (x y^{2} + 4 y) (y^{2} + 2 x y \frac{d y}{d x} + 4 \frac{d y}{d x})

f (x) = \frac{1}{2 x ^{3}}

\frac{d}{d x} (ln (\frac{x}{x ^{2} + 5}))

\int - \frac{2}{t + 1} d t

9 x - 4 y x^{2} + 1 \frac{d y}{d x} = 0, y (0) = 2

f (x) = 5 - 2 x^{2}, at x = - 2

What is the derivative of 1/5 (xy^2+4y^2) ?
The derivative of 1/5 (xy^2+4y^2) is 2/5 (xy^2+4y)(y^2+2xy(dy)/(dx)+4(dy)/(dx))
What is the first derivative of 1/5 (xy^2+4y^2) ?
The first derivative of 1/5 (xy^2+4y^2) is 2/5 (xy^2+4y)(y^2+2xy(dy)/(dx)+4(dy)/(dx))