y = 6 sin (4 TTx) (TT = pi)
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$\displaystyle y = k\sin(u)$ , where $\displaystyle u$ is a function of $\displaystyle x$ ... $\displaystyle \frac{dy}{dx} = k\cos(u) \cdot \frac{du}{dx}$
dy/dx= (6 cos(4TTx))(4TT) ?
Originally Posted by tradar dy/dx= (6 cos(4TTx))(4TT) ? $\displaystyle \frac{dy}{dx} = 24\pi \cos(4\pi x)$ it's ok to clean up the algebra.
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