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wcaprar
Find the second derivative of the parametric equation.
Parametric Equation: x=9t, y=6t-8 I got dy/dx = \[4/((1/2)t ^{-1/2})\] Now I need to find d^2y/dx^2. And I got \[-t ^{-3/2}/((1/2)t ^{-1/2})^3\] But It says that I got it wrong. Where did I go wrong?
wouldn't dy/dx = 2/3 \[\frac{dy}{dx} = \frac{dy}{dt}*\frac{dt}{dx} = 6*\frac{1}{9} = \frac{2}{3}\]
I would have used substitution t = x/9 then y = 6(x/9) - 8 dy/dx = 2/3 then 2nd derivative d^y/dx^2 = 0
How did you get: dy/dx = 4/((1/2)t−1/2)?