lf $f (x) = {sin}^{6} x + {cos}^{6} x$ then $f^{''} (x) =$

- 6 sin 4 x

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

6 sin 4 x

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

- 6 cos 4 x

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

6 cos 4 x

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

Open in App

Solution

The correct option is C $- 6 cos 4 x$
$f^{'} (x) = 6 {sin}^{5} x cos x + 6 {cos}^{5} x (- sin x)$
$= 3 sin 2 x ({sin}^{4} x - {cos}^{4} x)$
$= - 3 sin 2 x cos 2 x$
$= - \frac{3}{2} sin 4 x$
$f^{''} (x) = - 6 cos 4 x$

Suggest Corrections

Similar questions

\frac{cos 6 x + 6 cos 4 x + 15 cos 2 x + 10}{cos 5 x + 5 cos 3 x + 10 cos x}

E = $\frac{c o s 6 x + 6 c o s 4 x + 15 c o s 2 x + 10}{(c o s 5 x + 5 c o s 3 x + 10 c o s x)}$ equals to

What is the value of ∆ if (cos6x + 6cos4x + 15cos2x + 10)/(cos5x + 5cos3x + 10cosx)= ∆cosx ?

6 {cos}^{4} x - 7 {sin}^{4} x

f (x) = {sin}^{6} x + {cos}^{6} x, x \in R,

f (x)

Join BYJU'S Learning Program