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

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Question

(C) Let $y = \sqrt{x^2 + 16}$ and $z = \frac{x}{x - 1}$.First,differentiate $y$ with respect to $x$:$\frac{dy}{dx} = \frac{1}{2\sqrt{x^2 + 16}} \cdot

Vedclass · Accepted Answer

$-\frac{12}{5}$

Vedclass · Answer

(C) Let $y = \sqrt{x^2 + 16}$ and $z = \frac{x}{x - 1}$.First,differentiate $y$ with respect to $x$:$\frac{dy}{dx} = \frac{1}{2\sqrt{x^2 + 16}} \cdot (2x) = \frac{x}{\sqrt{x^2 + 16}}$.Next,differentiate $z$ with respect to $x$ using the quotient rule:$\frac{dz}{dx} = \frac{(x - 1)(1) - x(1)}{(x - 1)^2} = \frac{-1}{(x - 1)^2}$.Now,find the rate of change of $y$ with respect to $z$:$\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{x}{\sqrt{x^2 + 16}} \cdot (-(x - 1)^2) = \frac{-x(x - 1)^2}{\sqrt{x^2 + 16}}$.Substitute $x = 3$ into the expression:$\left( \frac{dy}{dz} \right)_{x = 3} = \frac{-3(3 - 1)^2}{\sqrt{3^2 + 16}} = \frac{-3(2^2)}{\sqrt{9 + 16}} = \frac{-3(4)}{\sqrt{25}} = \frac{-12}{5}$.

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

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