Can someone explain how you step by step find the slope to questions like this along with the answer? I forget how but I also ca
n’t figure it out and I’d like to know Incase I have more questions like this
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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)