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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Vedclass · Accepted Answer

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

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

Similar Questions

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then ${\left( {\frac{{dy}}{{dx}}} \right)^2}$ equals

If $x=a(t-\sin t)$ and $y=a(1+\cos t)$, then $\frac{d^2 y}{d x^2}=$

If $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$,then $\sqrt{1 + \left( \frac{dy}{dx} \right)^2} = $

If $x = at^2$ and $y = 2at$,then $\frac{dy}{dx} =$ . . . . . . ,where $t \neq 0$.

If $x=a \cos^{3} \theta$ and $y=a \sin^{3} \theta$,then $1+\left(\frac{dy}{dx}\right)^{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module