Derivatives of Trigonometric Functions: Solving for dy/dx and d²y/dx²

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SUMMARY

The discussion focuses on finding the first and second derivatives, dy/dx and d²y/dx², for the parametric equations x = cos(3t) and y = (sin(3t))². The derivatives are calculated using the chain rule, where dy/dx is derived from the ratio (dy/dt)/(dx/dt). The specific derivatives are dx/dt = -3sin(3t) and dy/dt = 6sin(3t)cos(3t), confirming the application of Leibniz notation in this context.

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Homework Statement


If x = cos(3t) and y = (sin(3t))² find dy/dx and d²y/dx².

Homework Equations


d/dx cosx = -sinx
d/dx sinx = cosx

The Attempt at a Solution


dx/dt = -3sin3t
dy/dt = 6sin3tcos3t

Not sure what to do now.
 
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(dy/dt)/(dx/dt)=dy/dx, right?
 
Yes, thank you. After I posted I wondered whether you could do that or not (a little unfamiliar with Leibniz notation) but it makes sense now thank you.
 

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