Given: x = e2t - 1 and y = e2t + 1. Find y’.
x=e2t−1x=e^{2t}-1x=e2t−1
=>dxdt=2e2t=> \frac{dx}{dt}=2e^{2t}=>dtdx=2e2t
y=e2t+1y=e^{2t}+1y=e2t+1
=>dydt=2e2t=> \frac{dy}{dt}=2e^{2t}=>dtdy=2e2t
y′=dydx=dydt×dtdxy'=\frac{dy}{dx}=\frac{dy}{dt}×\frac{dt}{dx}y′=dxdy=dtdy×dxdt
=>y′=dydx=2e2t1×12e2t=> y'=\frac{dy}{dx}=\frac{2e^{2t}}{1}×\frac{1}{2e^{2t}}=>y′=dxdy=12e2t×2e2t1
=>y′=dydx=1=> y'=\frac{dy}{dx}=1=>y′=dxdy=1
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