Mathematics, 29.10.2019 00:31 ani61
Find dy/dx and d2y/dx2. x = t2 + 4, y = t2 + 3t dy dx = d2y dx2 = for which values of t is the curve concave upward? (enter your answer using interval notation.)
Answers: 2
. Then by the chain rule,![\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{\mathrm dy}{\mathrm dx}\right]=\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dt}\cdot\dfrac{\mathrm dt}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}](/tpl/images/0350/2604/f6d89.png)
![\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac{\mathrm d}{\mathrm dt}\left[\frac{3t^2+3}{2t}\right]}{2t}=\dfrac{3(t^2-1)}{4t^3}](/tpl/images/0350/2604/ad5ec.png)
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Find dy/dx and d2y/dx2. x = t2 + 4, y = t2 + 3t dy dx = d2y dx2 = for which values of t is the curve...
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