Find the rate of change of sqrt{x^2 + 16} wit | vedclass

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(A) Let $u = \sqrt{x^2 + 16}$ and $v = \frac{x}{x - 1}$.First,differentiate $u$ with respect to $x$:$\frac{du}{dx} = \frac{1}{2\sqrt{x^2 + 16}} \cdot

Vedclass · Accepted Answer

$-12/5$

Vedclass · Answer

(A) Let $u = \sqrt{x^2 + 16}$ and $v = \frac{x}{x - 1}$.First,differentiate $u$ with respect to $x$:$\frac{du}{dx} = \frac{1}{2\sqrt{x^2 + 16}} \cdot 2x = \frac{x}{\sqrt{x^2 + 16}}$.Next,differentiate $v$ with respect to $x$ using the quotient rule:$\frac{dv}{dx} = \frac{(x - 1)(1) - x(1)}{(x - 1)^2} = \frac{-1}{(x - 1)^2}$.Now,find the rate of change of $u$ with respect to $v$:$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{x}{\sqrt{x^2 + 16}} \cdot \frac{(x - 1)^2}{-1}$.Substitute $x = 3$ into the expression:$\left( \frac{du}{dv} \right)_{x=3} = \frac{3}{\sqrt{3^2 + 16}} \cdot \frac{(3 - 1)^2}{-1} = \frac{3}{\sqrt{25}} \cdot \frac{4}{-1} = \frac{3}{5} \cdot (-4) = -\frac{12}{5}$.

Find the rate of change of $\sqrt{x^2 + 16}$ with respect to $\frac{x}{x - 1}$ at $x = 3$.

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