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+26402

+5

Consider the function \(f(x) =\frac{4 \sin{(x)}-x\cdot \cos{(x)}-2x}{2+\cos{(x)} } \quad \text{ for }\quad 0\le x \le2\pi\)

1) Show that f'(x)=4cosx-cos^2x/(2+cosx)^2

\(\begin{array}{rcll} f(x) = \frac{4 \sin{(x)}-x\cdot \cos{(x)}-2x}{2+\cos{(x)} } &=& \frac{4 \sin{(x)}-x\cdot [ 2+ \cos{(x)} ] }{2+\cos{(x)} }\\ &=& \frac{4 \sin{(x)} } {2+\cos{(x)} } - \frac{x\cdot [ 2+ \cos{(x)} ] } {2+\cos{(x)} }\\ &=& \frac{4 \sin{(x)} } {2+\cos{(x)} } - x\\\\ f(x) &=& \frac{u}{v} \qquad \Rightarrow \qquad f'(x) = \frac{u'v-uv'}{v^2} \\ u= 4 \sin{(x)} && u' = 4 \cos{(x)} \\ v= 2+\cos{(x)} && v' = - \sin{(x)} \\\\ f'(x) &=& \frac{ 4 \cos{(x)}(2+\cos{(x)}) - 4 \sin{(x)}(- \sin{(x)}) } {[2+\cos{(x)}]^2} - 1 \\ f'(x) &=& \frac{ 8 \cos{(x)} + 4 \cos^2{(x)} + 4 \sin^2{(x)} } {[2+\cos{(x)}]^2} - 1 \\ f'(x) &=& \frac{ 8 \cos{(x)} + 4 [\cos^2{(x)} + \sin^2{(x)}] } {[2+\cos{(x)}]^2} - 1 \qquad & \cos^2{(x)} + \sin^2{(x)} = 1\\ f'(x) &=& \frac{ 8 \cos{(x)} + 4 } {[2+\cos{(x)}]^2} - 1 \\ f'(x) &=& \frac{ 8 \cos{(x)} + 4 - [2+\cos{(x)}]^2} {[2+\cos{(x)}]^2} \\ f'(x) &=& \frac{ 8 \cos{(x)} + 4 - [4 + 4\cos{(x)} + \cos^2{(x)}]} {[2+\cos{(x)}]^2} \\ f'(x) &=& \frac{ 8 \cos{(x)} + 4 - 4 - 4\cos{(x)} - \cos^2{(x)}]} {[2+\cos{(x)}]^2} \\ f'(x) &=& \frac{ 8 \cos{(x)} - 4\cos{(x)} - \cos^2{(x)} } {[2+\cos{(x)}]^2} \\ f'(x) &=& \frac{ 4 \cos{(x)} - \cos^2{(x)}} {[2+\cos{(x)}]^2} \\ \end{array}\)

2.

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heurekaFeb 23, 2016

#8

+1023

0

SAVAGE!

Olivia3.141592654Feb 23, 2016

#1

+1023

0

I have it fourth hour.

Olivia3.141592654Feb 23, 2016

#11

+1023

0

Ya.

Olivia3.141592654Feb 23, 2016

#1

+1023

0

It's the quadratic formula. What do you need help with?

Olivia3.141592654Feb 23, 2016

#10

+263

0

Tru

FAFA11032000Feb 23, 2016

#3

+148

+5

okay.

Number Repeating Cycle of Sum of Digits of Multiples

2 {2,4,6,8,1,3,5,7,9}

3 {3,6,9,3,6,9,3,6,9}

4 {4,8,3,7,2,6,1,5,9}

5 {5,1,6,2,7,3,8,4,9}

6 {6,3,9,6,3,9,6,3,9}

7 {7,5,3,1,8,6,4,2,9}

8 {8,7,6,5,4,3,2,1,9}

9 {9,9,9,9,9,9,9,9,9}

10 {1,2,3,4,5,6,7,8,9}

11 {2,4,6,8,1,3,5,7,9}

12 {3,6,9,3,6,9,3,6,9}

13 {4,8,3,7,2,6,1,5,9}

It is asserted that the sum of digits follows a repeating sequence of length 9. This is so because for any decimal representation x, the number 10*x (ten times x) will have the same sum of digits. Multiplication by the base number 10 simply moves the digits one place to the left and puts a zero in the units place. More generally,

InstagramModelFeb 23, 2016

#9

+1023

0

Boring!

Olivia3.141592654Feb 23, 2016

#2