Answer to Question #301872 in Calculus for kazi

Question

Given: x = e^2t - 1 and y = e^2t + 1. Find y’.

Answer 1

$x=e^{2t}-1$

$=> \frac{dx}{dt}=2e^{2t}$

$y=e^{2t}+1$

$=> \frac{dy}{dt}=2e^{2t}$

$y'=\frac{dy}{dx}=\frac{dy}{dt}×\frac{dt}{dx}$

$=> y'=\frac{dy}{dx}=\frac{2e^{2t}}{1}×\frac{1}{2e^{2t}}$

$=> y'=\frac{dy}{dx}=1$