The rate of change of \( \sqrt{x^2-2} \) with respect to \( \displaystyle\frac{x}{x+1} \) at \( x=2 \) is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 9\sqrt{2} \)
Let \( u=\sqrt{x^2-2} \) and \( v=\frac{x}{x+1} \). The rate of change of u with respect to v is \( \frac{du}{dv}=\frac{du/dx}{dv/dx} \).
By the chain rule and the quotient rule:
\( \displaystyle \frac{du}{dx}=\frac{2x}{2\sqrt{x^2-2}}=\frac{x}{\sqrt{x^2-2}} \)
\( \displaystyle \frac{dv}{dx}=\frac{(x+1)-x}{(x+1)^2}=\frac{1}{(x+1)^2} \)
At x = 2:
\( \displaystyle \frac{du}{dx}=\frac{2}{\sqrt{2}}=\sqrt{2} \)
\( \displaystyle \frac{dv}{dx}=\frac{1}{9} \)
\( \displaystyle \frac{du}{dv}=\frac{\sqrt{2}}{1/9}=9\sqrt{2} \)