Find the nth derivative of (cosx)^9
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Answer:
y ' = 9 * cos^8(x) * (-sin x) = -9* cos^8(x)*sin(x)
y'' = -9*cos^8(x) * cos(x) + 72*cos^7(x)*sin^2(x)
= 72*cos^7(x)*sin^2(x) - 9*cos^9(x) <-------------- 1st term: 9 * 8 = 72 ; the exponents add up to 9
[done with calculator]
y''' = 255*cos^8(x)*sin(x) - 504*cos^6(x)*sin^3(x) <--- 2nd term: 7 * 72 = 504; exponents add up to 9
[done with calculator]
y(4) = 3024* cos^5(x)*sin^4(x) - 3122 * cos^7(x)*sin^2(x) + 255*cos^9(x) <--- 504*6 = 3024; expo. add up to 9
So in the next derivative it shall be 3024*5 = 15120*cos^4(x)*sin^5(x)
you then need to find a pattern for the other terms in the series
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