A ball is kicked 4 feet above the ground with an initial vertical velocity of 51 feet per second. The function h(t)=−16t²+51t+4
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One thing to bear in mind with trigonometric expressions is that, the exponent location is next to the function name, instead of the whole expression, though it applies to the whole thing, namely cos²(x), is really [ cos(x) ]², and that matters for using the chain rule.
![\bf y=cos^2(x)-sin^2(x)\implies y=[cos(x)]^2-[sin(x)]^2 \\\\\\ \cfrac{dy}{dx}=\stackrel{chain~rule}{2cos(x)\cdot -sin(x)}~~-~~\stackrel{chain~rule}{2sin(x)\cdot cos(x)} \\\\\\ \cfrac{dy}{dx}=-2cos(x)sin(x)-2cos(x)sin(x) \\\\\\ \cfrac{dy}{dx}=-4cos(x)sin(x)](https://tex.z-dn.net/?f=%5Cbf%20y%3Dcos%5E2%28x%29-sin%5E2%28x%29%5Cimplies%20y%3D%5Bcos%28x%29%5D%5E2-%5Bsin%28x%29%5D%5E2%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D%5Cstackrel%7Bchain~rule%7D%7B2cos%28x%29%5Ccdot%20-sin%28x%29%7D~~-~~%5Cstackrel%7Bchain~rule%7D%7B2sin%28x%29%5Ccdot%20cos%28x%29%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D-2cos%28x%29sin%28x%29-2cos%28x%29sin%28x%29%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D-4cos%28x%29sin%28x%29)