# Thread: Mathematics <Derivatives of Sin>

1. ## Mathematics <Derivatives of Sin>

I have a problem that doesn't make sense to me.
I will post the question, then the work I have completed on it so far after.

Question:
Find the slope of the tangent line to the sine function at the origin. Compare this value with the number of complete cycles in the interval [0, 2p]. What can you conclude about the slope of the sine function sin(ax) at the origin?

f(x)=sin(x)
g(x)=sin(2x)

my graphing calculator shows a deffinate steeper incline at the origin. This would make me assume that the Δy/Δx would be increasing as the number of cycles increase, but my calculator shows me different when calculating the the slope from the derivatives of these functions.

f(x)=sin(x)
f'(x)=cos(x)

g(x)=sin(2x)
g'(x)=cos(2x)

when i evaluated f'(0) and g'(0) I get 1 for each of them. if the slope is visually identifiable as steeper, does that not mean the slope would also be analytically steeper? i know the domain for sin and cos are [-1,1], so i'm assuming this would reconfirm the calculators information, but the sloops do not look the same at point (0,0).

Please if anyone can see where I'm confused or going wrong, point me to the light.

2. g(x)=sin(2x)
g'(x)=cos(2x)

Actually, g'(x) = 2cos(2x).

3. thanks for the input.
i just noticed i was using the chain rule incorrectly on a few other problems also.
that fixes it.
;0)

anyway I have another question on the chain rule.

knowing:
dy/dx[f(u)] = f'(u)u'
y=cotx
y'=-csc^2(x)
z=cscx
z'=-cscxcotx
sinx=(1/cscx)

then shouldn't this be correct?
f(x)=cotx/sinx
f'(x)=-csc^2(x)(-cscxcotx)
=csc^3(x)cotx
=cotx/sin^3x

the book says
[-1-cos^2(x)]/sin^3(x)

i'm assuming the book wants the answer in sin / cos.
how would i get the [-1-cos^2(x)]
i'm not aware of any trig convertion that would create this.

4. Just convert it into sinuses and cosinuses and derive:

f(x)=cotx/sinx = cos(x) / sin^2(x)

f'(x) = [division rule] = (-sin^3(x) - 2sin(x)cos^2(x)) / sin^4(x) = (-1 - 2cos^2(x)) / sin^3(x)
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