Differentiate:

Solution:

To find: Differentiation of $x^{3} \sec x$

Formula used: (i) (uv)' $=u^{\prime} v+u v^{\prime}$ (Leibnitz or product rule)

(ii)

$\frac{d x^{n}}{d x}=n x^{n-1}$

(iii)

$\frac{d \sec x}{d x}=\sec x \tan x$

Let us take $u=x^{3}$ and $v=\sec x$

$u^{\prime}=\frac{d u}{d x}=\frac{d x^{3}}{d x}=3 x^{2}$

$v^{\prime}=\frac{d v}{d x}=\frac{d s e c x}{d x}=\sec x \tan x$

Putting the above obtained values in the formula :-

$(u v)^{\prime}=u^{\prime} v+u v^{\prime}$

$\left(x^{3} \sec x\right)^{\prime}=3 x^{2} x \sec x+x^{3} x \sec x \tan x$

$=3 x^{2} \sec x+x^{3} \sec x \tan x$

$=x^{2} \sec x(3+x \tan x)$

Ans) $x^{2} \sec x(3+x \tan x)$