Trig Functions question

[Anthonyk2013]

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

 

[srmichael]

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

Not sure why you even have that 3rd step. From the second step, the derivative produces the last step.

 

[Anthonyk2013]

Not sure why you even have that 3rd step. From the second step, the derivative produces the last step.

Not the question I asked sr, that's the way it was written up on the black board. I don't understand how the sign changes?

 

[srmichael]

Not the question I asked sr, that's the way it was written up on the black board. I don't understand how the sign changes?

Whoever wrote it on the board was on something. The only thing I can guess is that since the derivative of cos is -sin, they took the minus from the -sin and used that to change the sign of the 2/3 fraction. Just not sure why they wrote it like that without the trig function accompanying it.

 

[Anthonyk2013]

Whoever wrote it on the board was on something. The only thing I can guess is that since the derivative of cos is -sin, they took the minus from the -sin and used that to change the sign of the 2/3 fraction. Just not sure why they wrote it like that without the trig function accompanying it.

Thanks I understand what your saying, notes seem to be missing some steps.

 

[JeffM]

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.

\(\displaystyle r = 5p \implies \dfrac{dr}{dp} = 5.\)

\(\displaystyle p = 2x \implies \dfrac{dp}{dx} = 2.\)

\(\displaystyle So\ \left(\dfrac{2}{3} * \dfrac{ds}{dx}\right) = \left(\dfrac{2}{3} * \dfrac{ds}{dr} * \dfrac{dr}{dp} * \dfrac{dp}{dx}\right) = \left(\dfrac{2}{3} * \{- sin(r)\} * 5 * 2\right) = \dfrac{20}{3} * \{- sin(5p)\} = \dfrac{20}{3} * \{- sin(10x)\} = \dfrac{-20}{3} * sin(10x).\)

\(\displaystyle Similarly,\ - \left(\dfrac{2}{5} * \dfrac{dq}{dx}\right) = \dfrac{-2}{5} * \dfrac{dq}{dp} * \dfrac{dp}{dx} = \dfrac{-2}{5} * cos(p) * 2 = \dfrac{-4}{5} * cos(2x).\)

\(\displaystyle THUS,\ \dfrac{dy}{dx} = \dfrac{-20sin(10x)}{3} + \dfrac{-4cos(2x)}{5}.\)

 

[Anthonyk2013]

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.

\(\displaystyle r = 5p \implies \dfrac{dr}{dp} = 5.\)

\(\displaystyle p = 2x \implies \dfrac{dp}{dx} = 2.\)

\(\displaystyle So\ \left(\dfrac{2}{3} * \dfrac{ds}{dx}\right) = \left(\dfrac{2}{3} * \dfrac{ds}{dr} * \dfrac{dr}{dp} * \dfrac{dp}{dx}\right) = \left(\dfrac{2}{3} * \{- sin(r)\} * 5 * 2\right) = \dfrac{20}{3} * \{- sin(5p)\} = \dfrac{20}{3} * \{- sin(10x)\} = \dfrac{-20}{3} * sin(10x).\)

\(\displaystyle Similarly,\ - \left(\dfrac{2}{5} * \dfrac{dq}{dx}\right) = \dfrac{-2}{5} * \dfrac{dq}{dp} * \dfrac{dp}{dx} = \dfrac{-2}{5} * cos(p) * 2 = \dfrac{-4}{5} * cos(2x).\)

\(\displaystyle THUS,\ \dfrac{dy}{dx} = \dfrac{-20sin(10x)}{3} + \dfrac{-4cos(2x)}{5}.\)

Thanks I appreciate your help. I have been away from the books for a good few years.