Trig Functions question

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Anthonyk2013

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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
 

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Anthonyk2013 said:
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
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Not sure why you even have that 3rd step. From the second step, the derivative produces the last step.
 
srmichael said:
Not sure why you even have that 3rd step. From the second step, the derivative produces the last step.
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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?
 
Anthonyk2013 said:
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?
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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.
 
srmichael said:
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.
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Thanks I understand what your saying, notes seem to be missing some steps.
 
Anthonyk2013 said:
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
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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.

y=2cos(10x)3−25∗sin(2x)=23∗cos(10x)−25∗sin(2x).\displaystyle y = \dfrac{2cos(10x)}{3} - \dfrac{2}{5} * sin(2x) = \dfrac{2}{3} * cos(10x) - \dfrac{2}{5} * sin(2x).

Let p=2x, q=sin(p), r=5p, and s=cos(r).\displaystyle Let\ p = 2x,\ q = sin(p),\ r = 5p,\ and\ s = cos(r).

So y=(23∗s)−(25∗q)  ⟹  dydx=(23∗dsdx)−(25∗dqdx).\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).

s=cos(r)  ⟹  dsdr=−sin(r).\displaystyle s = cos(r) \implies \dfrac{ds}{dr} = - sin(r). NOTICE the minus sign on the derivative.

r=5p  ⟹  drdp=5.\displaystyle r = 5p \implies \dfrac{dr}{dp} = 5.

p=2x  ⟹  dpdx=2.\displaystyle p = 2x \implies \dfrac{dp}{dx} = 2.

So (23∗dsdx)=(23∗dsdr∗drdp∗dpdx)=(23∗{−sin(r)}∗5∗2)=203∗{−sin(5p)}=203∗{−sin(10x)}=−203∗sin(10x).\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).

Similarly, −(25∗dqdx)=−25∗dqdp∗dpdx=−25∗cos(p)∗2=−45∗cos(2x).\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).

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

y=2cos(10x)3−25∗sin(2x)=23∗cos(10x)−25∗sin(2x).\displaystyle y = \dfrac{2cos(10x)}{3} - \dfrac{2}{5} * sin(2x) = \dfrac{2}{3} * cos(10x) - \dfrac{2}{5} * sin(2x).

Let p=2x, q=sin(p), r=5p, and s=cos(r).\displaystyle Let\ p = 2x,\ q = sin(p),\ r = 5p,\ and\ s = cos(r).

So y=(23∗s)−(25∗q)  ⟹  dydx=(23∗dsdx)−(25∗dqdx).\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).

s=cos(r)  ⟹  dsdr=−sin(r).\displaystyle s = cos(r) \implies \dfrac{ds}{dr} = - sin(r). NOTICE the minus sign on the derivative.

r=5p  ⟹  drdp=5.\displaystyle r = 5p \implies \dfrac{dr}{dp} = 5.

p=2x  ⟹  dpdx=2.\displaystyle p = 2x \implies \dfrac{dp}{dx} = 2.

So (23∗dsdx)=(23∗dsdr∗drdp∗dpdx)=(23∗{−sin(r)}∗5∗2)=203∗{−sin(5p)}=203∗{−sin(10x)}=−203∗sin(10x).\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).

Similarly, −(25∗dqdx)=−25∗dqdp∗dpdx=−25∗cos(p)∗2=−45∗cos(2x).\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).

THUS, dydx=−20sin(10x)3+−4cos(2x)5.\displaystyle THUS,\ \dfrac{dy}{dx} = \dfrac{-20sin(10x)}{3} + \dfrac{-4cos(2x)}{5}.
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Thanks I appreciate your help. I have been away from the books for a good few years.
 
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