SOLUTION: Can anyone please help?!
Determine the following derivatives with respect to theta (th):
i) y = ln (2 pi th)
ii) y = Sin^-1 (2th)
iii) For the curve y = th^3 - 6th^2 +
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Question 78356: Can anyone please help?!
Determine the following derivatives with respect to theta (th):
i) y = ln (2 pi th)
ii) y = Sin^-1 (2th)
iii) For the curve y = th^3 - 6th^2 + 9th + 24 determine the coordinates of the turning points and state the condition of each turning point.
Thanks!
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Remember the derivative of the natural log is 1/x so we get
=\frac{1}{2\pi\theta})
Now we derive the inside since we must follow the chain rule:
=\frac{1}{2\pi\theta}*\frac{d}{d\theta}(2\pi\theta))
So we eventually get:
=\frac{1}{2\pi\theta}*\frac{d}{d\theta}(2\pi\theta)=\frac{2\pi}{2\pi\theta})
Which reduces to:
Remember the derivative of arcsine is 
So we get
)=\frac{1}{sqrt{1-(2\theta)^2}}=\frac{2}{sqrt{1-4\theta^2}})
Remember to derive 2 theta as well (by using the chain rule)
If you were to take the derivative of x^3 - 6x^2 + 9x + 24 and set it equal to zero, then you'd find your turning points.
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