Find the 80th derivative of f(c)
ET
sin(2).
f(80)
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Compute the first few derivatives to reveal a pattern:

f(x) = x sin(x)

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

f ''(x) = (-x sin(x) + cos(x)) + cos(x) = -x sin(x) + 2 cos(x)

f '''(x) = (-x cos(x) - sin(x)) - 2 sin(x) = -x cos(x) - 3 sin(x)

f (x) = (x sin(x) - cos(x)) - 3 cos(x)) = x sin(x) - 4 cos(x)

f ⁽⁵⁾(x) = (x cos(x) + sin(x)) + 4 sin(x) = x cos(x) + 5 sin(x)

f ⁽⁶⁾(x) = (-x sin(x) + cos(x)) + 5 cos(x) = -x sin(x) + 6 cos(x)

f ⁽⁷⁾(x) = (-x cos(x) - sin(x)) - 6 sin(x) = -x cos(x) - 7 sin(x)

f ⁽⁸⁾(x) = (x sin(x) - cos(x)) - 7 cos(x) = x sin(x) - 8 cos(x)

and so on.

The pattern should be clear: for some integer k,

• if n = 4k, then f ⁽ⁿ ⁾(x) = x sin(x) - n cos(x)

• if n = 4k + 1, then f ⁽ⁿ ⁾(x) = x cos(x) + n sin(x)

• if n = 4k + 2, then f ⁽ⁿ ⁾(x) = -x sin(x) + n cos(x)

• if n = 4k + 3, then f ⁽ⁿ ⁾(x) = -x cos(x) - n cos(x)

80 is a multiple of 4, so the 80th derivative falls in the first category. Then

f ⁽⁸⁰⁾(x) = x sin(x) - 80 cos(x)

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2