Question #301872

Given: x = e2t - 1 and y = e2t + 1. Find y’.


1
Expert's answer
2022-02-27T12:20:03-0500

x=e2t1x=e^{2t}-1


=>dxdt=2e2t=> \frac{dx}{dt}=2e^{2t}


y=e2t+1y=e^{2t}+1


=>dydt=2e2t=> \frac{dy}{dt}=2e^{2t}


y=dydx=dydt×dtdxy'=\frac{dy}{dx}=\frac{dy}{dt}×\frac{dt}{dx}


=>y=dydx=2e2t1×12e2t=> y'=\frac{dy}{dx}=\frac{2e^{2t}}{1}×\frac{1}{2e^{2t}}



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



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