How do you find the nth derivative of f(x) = sin^4(x) + sin^2(x)cos^2(x)?

Aug 14, 2015

${f}^{n} \left(x\right) = {2}^{n - 1} \left({\delta}_{0} + \cos \left(2 x + \frac{\left(n - 2\right) \pi}{2}\right)\right) \text{ } n \in {\mathbb{Z}}_{0}^{+}$

where ${\delta}_{0} = \left\{\text{_(0, " otherwise")^( 1, " } n = 0\right)$

Explanation:

${\sin}^{2} A + {\cos}^{2} A = 1$
$\sin A = \cos \left(A - \frac{\pi}{2}\right)$
$\cos 2 A = 1 - 2 {\sin}^{2} A$

$f \left(x\right) = {\sin}^{4} x + {\sin}^{2} x {\cos}^{2} x = {\sin}^{2} x \left({\sin}^{2} x + {\cos}^{2} x\right) = \frac{1}{2} \left(1 - \cos 2 x\right) = \frac{1}{2} \left(1 + \cos \left(2 x - \pi\right)\right)$

$f ' \left(x\right) = \sin 2 x = \cos \left(2 x - \frac{\pi}{2}\right)$
$f ' ' \left(x\right) = 2 \cos 2 x = 2 \cos \left(2 x + 0 \pi\right)$
$f ' ' ' \left(x\right) = - 4 \sin 2 x = - 4 \cos \left(2 x - \frac{\pi}{2}\right) = 4 \cos \left(2 x + \frac{\pi}{2}\right)$
$f ' ' ' ' \left(x\right) = - 8 \cos 2 x = 8 \cos \left(2 x + \pi\right)$

${f}^{n} \left(x\right) = {2}^{n - 1} \left({\delta}_{0} + \cos \left(2 x + \frac{\left(n - 2\right) \pi}{2}\right)\right) \text{ } n \in {\mathbb{Z}}_{0}^{+}$

where ${\delta}_{0} = \left\{\text{_(0, " otherwise")^( 1, " } n = 0\right)$