Did this question last night in class, Not sure how we changed the 2 circled from a positive 2 to a negative 2. Any ideas
Like Sir Michael, I am confused about the third line. It does not follow from the second line, and the fourth line does not follow from the third line. Moreover, using x to stand for multiplication in a problem involving x as a variable is something you are taught not to do in seventh or eighth grade. Nevertheless, I think what is going on is this.
\(\displaystyle y = \dfrac{2cos(10x)}{3} - \dfrac{2}{5} * sin(2x) = \dfrac{2}{3} * cos(10x) - \dfrac{2}{5} * sin(2x).\)
\(\displaystyle Let\ p = 2x,\ q = sin(p),\ r = 5p,\ and\ s = cos(r).\)
\(\displaystyle So\ y = \left(\dfrac{2}{3} * s\right) - \left(\dfrac{2}{5} * q\right) \implies \dfrac{dy}{dx} = \left(\dfrac{2}{3} * \dfrac{ds}{dx}\right) - \left(\dfrac{2}{5} * \dfrac{dq}{dx}\right).\)
\(\displaystyle s = cos(r) \implies \dfrac{ds}{dr} = - sin(r).\) NOTICE the minus sign on the derivative.
\(\displaystyle r = 5p \implies \dfrac{dr}{dp} = 5.\)
\(\displaystyle p = 2x \implies \dfrac{dp}{dx} = 2.\)
\(\displaystyle So\ \left(\dfrac{2}{3} * \dfrac{ds}{dx}\right) = \left(\dfrac{2}{3} * \dfrac{ds}{dr} * \dfrac{dr}{dp} * \dfrac{dp}{dx}\right) = \left(\dfrac{2}{3} * \{- sin(r)\} * 5 * 2\right) = \dfrac{20}{3} * \{- sin(5p)\} = \dfrac{20}{3} * \{- sin(10x)\} = \dfrac{-20}{3} * sin(10x).\)
\(\displaystyle Similarly,\ - \left(\dfrac{2}{5} * \dfrac{dq}{dx}\right) = \dfrac{-2}{5} * \dfrac{dq}{dp} * \dfrac{dp}{dx} = \dfrac{-2}{5} * cos(p) * 2 = \dfrac{-4}{5} * cos(2x).\)
\(\displaystyle THUS,\ \dfrac{dy}{dx} = \dfrac{-20sin(10x)}{3} + \dfrac{-4cos(2x)}{5}.\)