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Find derivatives of other trigonometric functions

Oirection:\newlineMath-AlgCBAv2222 (copy)\newline2828\newlineAlicia completed the square in the quadratic function p(x)=−2x2+24x−67 p(x)=-2 x^{2}+24 x-67 in order to determine the maximum value of p(x) p(x) . The new form of the function that resulted from her work was p(x)=−2(x−6)2+□ p(x)=-2(x-6)^{2}+\square \newlineWhat number should go in the box so that Alicia's new form is equivalent to the original form of p(x) p(x) ? Enter the answer as an integer or a decimal in the box.\newlinep(x)=−2(x−6)2+□ p(x)=-2(x-6)^{2}+\square
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Find the derivative of \newlinef(x)=6sec⁡(x)+x3.f(x)=6\sec(x)+x^{3}.
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Question\newline2424 Calculate the derivative dydx \frac{d y}{d x} for the following parametrically defined plane curve and locate any critical points on its graph.\newlinex(t)=2sin⁡(2t),y(t)=2cos⁡(t),0≤t≤2π. x(t)=2 \sin (2 t), \quad y(t)=2 \cos (t), \quad 0 \leq t \leq 2 \pi . \newlineSelect the correct answer below:\newlinedydx=−sin⁡(2t)cos⁡(t) \frac{d y}{d x}=-\frac{\sin (2 t)}{\cos (t)} , critical points at (0,±2),(±2,±2) (0, \pm 2),( \pm 2, \pm \sqrt{2}) \newlinedydx=−sin⁡(t)cos⁡(2t) \frac{d y}{d x}=-\frac{\sin (t)}{\cos (2 t)} , critical points at (0,±2),(±2,±2) (0, \pm \sqrt{2}),( \pm 2, \pm \sqrt{2}) \newlinedydx=−sin⁡(t)2cos⁡(2t) \frac{d y}{d x}=-\frac{\sin (t)}{2 \cos (2 t)} , critical points at (0,±2),(±2,±2) (0, \pm 2),( \pm 2, \pm \sqrt{2}) \newlinedydx=−sin⁡(t)cos⁡(t) \frac{d y}{d x}=-\frac{\sin (t)}{\cos (t)} , critical points at (0,±2),(±2,±2) (0, \pm 2),( \pm \sqrt{2}, \pm 2) \newlineFEEDBACK\newlineMORE INSTRUCTION
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