The rate of change of sqrt{x^2+16} with respe | vedclass

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Question

(A) Let $y = \sqrt{x^2+16}$.Then,$\frac{dy}{dx} = \frac{1}{2\sqrt{x^2+16}} \cdot (2x) = \frac{x}{\sqrt{x^2+16}}$.Let $z = \frac{x}{x-1}$.Using the quo

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$\frac{-80}{\sqrt{41}}$

Vedclass · Answer

(A) Let $y = \sqrt{x^2+16}$.Then,$\frac{dy}{dx} = \frac{1}{2\sqrt{x^2+16}} \cdot (2x) = \frac{x}{\sqrt{x^2+16}}$.Let $z = \frac{x}{x-1}$.Using the quotient rule,$\frac{dz}{dx} = \frac{(x-1)(1) - x(1)}{(x-1)^2} = \frac{-1}{(x-1)^2}$.We need to find $\frac{dy}{dz} = \frac{dy/dx}{dz/dx}$.$\frac{dy}{dz} = \frac{x}{\sqrt{x^2+16}} \div \left( \frac{-1}{(x-1)^2} \right) = -\frac{x(x-1)^2}{\sqrt{x^2+16}}$.At $x=5$:$\frac{dy}{dz} = -\frac{5(5-1)^2}{\sqrt{5^2+16}} = -\frac{5(16)}{\sqrt{25+16}} = -\frac{80}{\sqrt{41}}$.

The rate of change of $\sqrt{x^2+16}$ with respect to $\frac{x}{x-1}$ at $x=5$ is

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