$n$ th derivative of $\cos^{2} (x) =$

$2^{n - 1} \cos (n \frac{π}{2} + 2 x)$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$2^{n} \cos (n \frac{π}{2} - 2 x)$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$2^{n + 1} \cos (n \frac{π}{2} + 2 x)$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$2^{n + 2} \cos (n \frac{π}{2} + 2 x)$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
$2^{n - 1} \cos (n \frac{π}{2} + 2 x)$

Finding the $n$ th derivative:
Let $y = \cos^{2} (x)$
Differentiate w.r.t. $x$
$y_{1} = - 2 \cos (x) \sin (x)$ $= - \sin (2 x)$
$y_{2} = - 2 \cos (2 x) y_{3} = 2^{2} \sin (2 x) y_{4} = 2^{3} \cos (2 x) y_{5} = - 2^{4} \sin (2 x)$
Similarly,
$y_{n} = 2^{n - 1} \cos (n \frac{π}{2} + 2 x)$
Hence, option(A) is the correct answer.

Suggest Corrections