Find the rate of change of the distance D between the origin and a mov

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Find the rate of change of the distance D between the origin and a moving point on the graph of y = x2 + 4 if dx dt = 2 centimeters per second. dD dt = 4 x3+10 xx4+4 x2+16 dD dt = 4 x3−18 xx3+9 x2+16 dD dt = 4 x3+18 xx4+9 x2+16 dD dt = 2 x3+18 xx4+8 x2+16 dD dt = 2 x3−10 xx3+9 x2+16

Find the rate of change of the distance D between the origin and a moving point on the graph of y = x2 + 4 if dx dt = 2 centimeters per second. dD dt = 4 x3+10 xx4+4 x2+16 dD dt = 4 x3−18 xx3+9 x2+16 dD dt = 4 x3+18 xx4+9 x2+16 dD dt = 2 x3+18 xx4+8 x2+16 dD dt = 2 x3−10 xx3+9 x2+16

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Find the rate of change of the distance

D

between the origin and a moving point on the graph of

y = x^{2} + 4

if

\frac{d x}{d t} = 2

centimeters per second.

\frac{d D}{d t} = \frac{4 x^{3} + 10 x}{x^{4} + 4 x^{2} + 16}

\frac{d D}{d t} = \frac{4 x^{3} - 18 x}{\sqrt{x^{3} + 9 x^{2} + 16}}

\frac{d D}{d t} = \frac{4 x^{3} + 18 x}{\sqrt{x^{4} + 9 x^{2} + 16}}

\frac{d D}{d t} = \frac{2 x^{3} + 18 x}{\sqrt{x^{4} + 8 x^{2} + 16}}

\frac{d D}{d t} = \frac{2 x^{3} - 10 x}{x^{3} + 9 x^{2} + 16}

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