InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Find \(\frac{dy}{dx}\), if x=6 sin^-12t and y=\(\frac{1}{\sqrt{1-4t^2}}\).(a) \(\frac{t}{1-4t^2}\)(b) –\(\frac{1}{3(1-4t^2)}\)(c) –\(\frac{t}{3(1-4t^2)}\)(d) \(\frac{1}{3(1-4t^2)}\)The question was asked during an internship interview.This interesting question is from Derivatives of Functions in Parametric Forms topic in section Continuity and Differentiability of Mathematics – Class 12 |
|
Answer» RIGHT answer is (d) \(\FRAC{1}{3(1-4T^2)}\) The explanation: Given that, x=6 sin^-12t and y=\(\frac{1}{\sqrt{1-4t^2}}\) \(\frac{DX}{dt}\)=\(\frac{6}{\sqrt{1-4t^2}}.2=\frac{12}{\sqrt{1-4t^2}}\) \(\frac{dy}{dt}\)=-\(\frac{1}{2(1-4t^2)^{3/2}}.(-8t)=\frac{4t}{(1-4t^2)^{3/2}}\) \(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=\frac{4t}{(1-4t^2)^{3/2}}.\frac{\sqrt{1-4t^2}}{12}\) \(\frac{dy}{dx}\)=\(\frac{t}{3(1-4t^2)}\) |
|