If #f(x)=sin2x# then what is the #75^(th)# derivative?

Question

Cameron Alexander · Accepted Answer

# f^((75))(x) = -2^75 cos2x #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -37778931862957161709568 cos2x #

Should we possess: # f(x) = sin2x # Then repeatedly differentiating wrt #x# we get: # \ \ \ \ f(x) = sin2x # # \ \ \ f'(x) = 2cos2x # # \ f''(x) = 2(-2sin2x) \ \ = -2^2 sin2x # # f'''(x) = -2^2 (2cos2x) = -2^3 cos2x # # f^((4))(x) = 2^4 sin2x # #vdots # After which the pattern continues with a cycle of 4, so we can represent the #n^(th)# derivative by: # f^((n))(x) = { (2^nsin2x, n mod 4 = 0), (2^ncos2x, n mod 4 = 1), (-2^nsin2x, n mod 4 = 2), (-2^ncos2x, n mod 4 = 3) :} # Now #75 mod 4 = 3#, and so we can deduce that: # f^((75))(x) = -2^75 cos2x # # \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -37778931862957161709568 cos2x #

Emma Aldridge · Answer

undefined To find the 75th derivative of $ f(x) = \sin(2x) $, we can use the fact that the derivative of $ \sin(ax) $ with respect to $ x $ is $ a\cos(ax) $. Repeatedly applying this rule, we see that the derivatives of $ f(x) $ will alternate between $ \sin(2x) $ and $ \cos(2x) $, with a factor of $ 2^n $ for every $ n $ derivatives taken.

Since the derivative of $ \sin(2x) $ is $ 2\cos(2x) $ and the derivative of $ \cos(2x) $ is $ -2\sin(2x) $, every even derivative will be of the form $ 2^n\sin(2x) $ and every odd derivative will be of the form $ 2^n\cos(2x) $.

For the 75th derivative:
- Since 75 is odd, the 75th derivative will be of the form $ 2^{37}\cos(2x) $.

Adam Adkins · Answer

undefined To find the 75th derivative of $ f(x) = \sin(2x) $, we can use the fact that the derivative of $ \sin(ax) $ with respect to $ x $ is $ a\cos(ax) $.

Given that $ f(x) = \sin(2x) $, the first derivative $ f'(x) $ would be $ \frac{d}{dx} \sin(2x) = 2\cos(2x) $.

Continuing to differentiate $ f(x) = \sin(2x) $ with respect to $ x $, the second derivative $ f''(x) $ would be $ \frac{d}{dx} (2\cos(2x)) = -4\sin(2x) $.

We can observe a pattern here: the derivatives of $ \sin(2x) $ alternate between $ \sin(2x) $ and $ \cos(2x) $, with the coefficient oscillating between $ 2 $ and $ -2 $ and the sine and cosine functions swapping places. This pattern repeats every second derivative.

Since the 75th derivative of $ \sin(2x) $ will be $ \sin(2x) $ if $ 75 $ is odd, and $ \cos(2x) $ if $ 75 $ is even, we need to determine the parity of $ 75 $.

$ 75 $ is an odd number, so the 75th derivative of $ \sin(2x) $ will be $ \sin(2x) $.