Question

differentiation of
$\sqrt{ \tan{}^{3}x }$
how to solve the
$\sqrt{ \tan{}^{3}x }$


differentiation of
​

Answer 1

Step-by-step explanation:

We have,

$\rm \: y = \sqrt{ \tan ^{3} (x) }$

Differentiating w.r.t. x,

$\rm \: \frac{dy}{dx} = \frac{1} {2\sqrt{ \tan ^{3} (x) }}. \frac{d}{dx} \{ \tan ^{3} (x) \}$

$\implies \rm \: \frac{dy}{dx} = \frac{1} {2\sqrt{ \tan ^{3} (x) }}. \{ 3\tan ^{2} (x). \sec ^{2} (x) \}$

$\implies \rm \: \frac{dy}{dx} = \frac{3\tan ^{2} (x). \sec ^{2} (x) } {2\sqrt{ \tan ^{3} (x) }}$

$\implies \rm \: \frac{dy}{dx} = \frac{3}{2} \frac{\tan ^{2} (x). \sec ^{2} (x) } {\tan ^{ 3 /2 } (x) }$

$\implies \rm \: \frac{dy}{dx} = \frac{3}{2} \tan ^{2 - \frac{3}{2} } (x). \sec ^{2} (x)$

$\implies \rm \: \frac{dy}{dx} = \frac{3}{2} \tan ^{ \frac{4 - 3}{2} } (x). \sec ^{2} (x)$

$\implies \rm \: \frac{dy}{dx} = \frac{3}{2} \tan ^{ \frac{1}{2} } (x). \sec ^{2} (x)$

$\implies \rm \: \frac{dy}{dx} = \frac{3}{2} \sqrt{\tan (x)}. \sec ^{2} (x)$