Question

99th derivative of sin2x

Answer 1

Answer:

So, 99th derivative is :  

f⁹⁹ = 2⁹⁹ cos(2x)

Explanation:

We handle the even and odd order derivatives apart.

f′(x) = 2cos(2x)                      1st derivative

f′′′(x) = −8cos(2x)                  2nd derivative

f′′′′′(x) = 32cos(2x)                3rd derivative

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$f^{2n+1}  = (-1)^{n} 2^{2n+1} cos(2x)$

Then

f(x) = sin(2x)

f′′(x) = −4sin(2x)

f′′′′(x) = 16sin(2x)

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$f^{2n}  = (-1)^{n} 2^{2n} sin(2x)$

Find 99th derivative. So, we plug n = 48 in $f^{2n+1}  = (-1)^{n} 2^{2n+1} cos(2x)$

$f^{2*48+1}  = (-1)^{48} 2^{2*48+1} cos(2x)\\f^{99}  = 2^{99} cos(2x)$

So, 99th derivative is :

f⁹⁹ = 2⁹⁹ cos(2x)